# The squeeze theorem – Serlo

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The squeeze theorem is a powerful tool to determine the limit of a complicated sequence. It is based on comparison to simpler sequences, for which the limit is easily determinable.

## Motivation

Convergence proof for a root sequence (video in German)

The intuition behind this theorem is quite simple: We are given a complicated sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ and want to know whether it converges. Often, one can leave out terms in the complicated sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ and gets some simpler sequences ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (c_{n})_{n\in \mathbb {N} }}$. If ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$ is a lower bond and ${\displaystyle (c_{n})_{n\in \mathbb {N} }}$ an upper bound, then ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is "caught" in the space between both functions. If both sequences converge to the same limit ${\displaystyle a}$, they "squeeze" together this space to this single point and ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ has no other option than to converge towards ${\displaystyle a}$, as well.

A ham & cheese sandwich

You may also visualize this theorem by a "ham & cheese sandwich" (see image on the right). The upper and lower bounds ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (c_{n})_{n\in \mathbb {N} }}$ act as two "slices of toast", which confine ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ , i.e. the "filling". If you squeeze the toast slices together, the filling in between will also squeezed to this point.

## The squeeze theorem

Squueze theorem (video in German, YouTube videofrom the YouTube channel Maths CA).

Theorem (Squeeze theorem)

Let ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ be any sequence. If two sequences ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (c_{n})_{n\in \mathbb {N} }}$ exist with ${\displaystyle b_{n}\leq a_{n}\leq c_{n}}$ for all ${\displaystyle n\in \mathbb {N} }$ and ${\displaystyle \lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }c_{n}=a}$ for some common limit ${\displaystyle a\in \mathbb {R} }$, then ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ will also converge towards this limit ${\displaystyle a}$.

Proof (Squeeze theorem)

Let ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (c_{n})_{n\in \mathbb {N} }}$ be such that ${\displaystyle b_{n}\leq a_{n}\leq c_{n}}$ and ${\displaystyle \lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }c_{n}=a}$ for some ${\displaystyle a\in \mathbb {R} }$.

We need to prove ${\displaystyle \lim _{n\to \infty }a_{n}=a}$ , i.e. for each ${\displaystyle \epsilon >0}$ there is an ${\displaystyle N\in \mathbb {N} }$, such that ${\displaystyle |a_{n}-a|<\epsilon }$ for all ${\displaystyle n\geq N}$ . Let ${\displaystyle \epsilon >0}$ be arbitrary. By assumption, the convergences ${\displaystyle \lim _{n\to \infty }b_{n}=a}$ and ${\displaystyle \lim _{n\to \infty }c_{n}=a}$ hold.

Hence, there are two thresholds ${\displaystyle N_{1}\in \mathbb {N} }$ and ${\displaystyle N_{2}\in \mathbb {N} }$ with ${\displaystyle |b_{n}-a|<\epsilon }$ for all ${\displaystyle n\geq N_{1}}$ (lower bounding sequence) and ${\displaystyle |c_{n}-a|<\epsilon }$ for all ${\displaystyle n\geq N_{2}}$ (upper bounding sequence). Now, there is ${\displaystyle b_{n}\leq a_{n}\leq c_{n}}$ for ${\displaystyle n\in \mathbb {N} }$. For each ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ we distinguish the two cases ${\displaystyle a_{n}\geq a}$ (above the limit) and ${\displaystyle a_{n}\leq a}$ (below the limit). In case ${\displaystyle a_{n}\geq a}$ there is

${\displaystyle |\underbrace {a_{n}-a} _{\geq 0}|=a_{n}-a{\overset {a_{n}\leq c_{n}}{\leq }}\underbrace {c_{n}-a} _{\geq 0}=|c_{n}-a|}$

The other case ${\displaystyle a_{n}\leq a}$ is treated analogously:

${\displaystyle |\underbrace {a_{n}-a} _{\leq 0}|=-(a_{n}-a)=a-a_{n}{\overset {a_{n}\geq b_{n}}{\leq }}a-b_{n}=-(\underbrace {b_{n}-a} _{\leq 0})=|b_{n}-a|}$

So if the maximum of both thresholds ${\displaystyle N=\max\{N_{1},N_{2}\}}$ is exceeded by ${\displaystyle n\geq N}$ ( i.e. ${\displaystyle n\geq N_{1}}$ and ${\displaystyle n\geq N_{2}}$ hold), then for ${\displaystyle a_{n}\geq a}$,

${\displaystyle |a_{n}-a|\leq |c_{n}-a|<\epsilon }$

and for ${\displaystyle a_{n}\leq a}$,

${\displaystyle |a_{n}-a|\leq |b_{n}-a|<\epsilon }$

So in either case ${\displaystyle |a_{n}-a|<\epsilon }$ for all ${\displaystyle n\geq N}$, which establishes convergence.

Hint

The inequality ${\displaystyle b_{n}\leq a_{n}\leq c_{n}}$ does not need to be satisfied for all ${\displaystyle n\in \mathbb {N} }$ . Assume, a finite amount of sequence elements ${\displaystyle a_{n}}$ is not "caught between ${\displaystyle b_{n}}$ and ${\displaystyle c_{n}}$". Then, after some finite ${\displaystyle n_{0}\in \mathbb {N} }$ all of these finite elements will have been passed and the further sequence elements ${\displaystyle n\geq n_{0}}$ will indeed be caught as ${\displaystyle b_{n}\leq a_{n}\leq c_{n}}$.

Example

We illustrate the squeeze theorem using the root sequence ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{c}}=1}$ for an arbitrary but fixed constant ${\displaystyle c>0}$.

Two cases need to be distinguished:

Fall 1: ${\displaystyle c\geq 1}$

Let us choose some ${\displaystyle n_{0}\in \mathbb {N} }$, such that ${\displaystyle n_{0}\geq c}$. Then, for all ${\displaystyle n\in \mathbb {N} }$ with ${\displaystyle n\geq n_{0}}$ there is ${\displaystyle c\leq n}$, so ${\displaystyle {\sqrt[{n}]{c}}\leq {\sqrt[{n}]{n}}}$, i.e. we have an upper bound. A lower bound is given by ${\displaystyle 1={\sqrt[{n}]{1}}\leq {\sqrt[{n}]{c}}}$ for any ${\displaystyle n\in \mathbb {N} }$.

In the previous chapter, we have established the limit ${\displaystyle \lim _{n\to \infty }1=1=\lim _{n\to \infty }{\sqrt[{n}]{n}}}$. So the squeeze theorem yields ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{c}}=1}$ for any ${\displaystyle c}$ standing inside the root.

Fall 2: ${\displaystyle c<1}$

In this case, ${\displaystyle {\tfrac {1}{c}}\geq 1}$, so we are led to the above case, replacing ${\displaystyle c}$ by ${\displaystyle {\tfrac {1}{c}}}$ . Hence, also ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{\tfrac {1}{c}}}=1}$ and the limit theorems (quotient) yield:

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{c}}=\lim _{n\to \infty }{\frac {1}{\sqrt[{n}]{\frac {1}{c}}}}={\frac {1}{\lim _{n\to \infty }{\sqrt[{n}]{c}}}}={\frac {1}{1}}=1}$

## A useful special case

We often encounter 0 as a sequence limit (null sequence). Since the squeeze theorem can be used to prove any ${\displaystyle a}$ to be a limit of a sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$, it can also be used for ${\displaystyle a=0}$. Especially, for any convergent ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$, the sequence ${\displaystyle (|a_{n}-a|)_{n\in \mathbb {N} }}$ must converge to 0:

Theorem (Special case of the squeeze theorem)

Let ${\displaystyle (c_{n})_{n\in \mathbb {N} }}$ be a null sequence and ${\displaystyle |a_{n}-a|\leq c_{n}}$ for all ${\displaystyle n\in \mathbb {N} }$. Then ${\displaystyle \lim _{n\to \infty }a_{n}=a}$.

Proof (Special case of the squeeze theorem)

We set ${\displaystyle {\tilde {a}}_{n}=|a_{n}-a|}$. For the constant sequence ${\displaystyle b_{n}=0}$ there is ${\displaystyle b_{n}\leq {\tilde {a_{n}}}\leq c_{n}}$ (absolute values are always positive). Since ${\displaystyle \lim _{n\to \infty }b_{n}=0=\lim _{n\to \infty }c_{n}}$ the squeeze theorem also implies ${\displaystyle \lim _{n\to \infty }{\tilde {a}}_{n}=0}$. Therefore, ${\displaystyle (|a_{n}-a|)_{n\in \mathbb {N} }}$ is a null sequence, which is equivalent to the statement that ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ converges to ${\displaystyle a}$ .

Hint

This special case of the squeeze theorem is also referred to as direct comparison argument for sequences .

Example

Let us take a look at ${\displaystyle \lim _{n\to \infty }{\sqrt {n^{2}+1}}-n=0}$. Not a simple case: both ${\displaystyle {\sqrt {n^{2}+1}}}$ and ${\displaystyle n}$ converge to ${\displaystyle \infty }$ . But we can show that its difference converges to 0, since it is bounded by:

${\displaystyle |{\sqrt {n^{2}+1}}-n|\leq {\frac {1}{2n}}}$

Let us prove this inequality. There is ${\displaystyle n^{2}+1>n^{2}}$ so ${\displaystyle {\sqrt {n^{2}+1}}>{\sqrt {n^{2}}}=n}$ and the difference is positive, i.e. ${\displaystyle {\sqrt {n^{2}+1}}-n>0}$ .

Now, we remove the root by squaring it:

{\displaystyle {\begin{aligned}&\ \ |{\sqrt {n^{2}+1}}-n|\leq {\frac {1}{2n}}\\[0.3em]\Leftrightarrow &\ \ {\sqrt {n^{2}+1}}\leq n+{\frac {1}{2n}}\\[0.3em]\Leftrightarrow &\ \ n^{2}+1\leq \left(n+{\frac {1}{2n}}\right)^{2}=n^{2}+1+{\frac {1}{4n^{2}}}\\[0.3em]\Leftrightarrow &\ \ 0\leq {\frac {1}{4n^{2}}}\end{aligned}}}

The last inequality holds for all ${\displaystyle n\in \mathbb {N} }$ and implies

${\displaystyle |{\sqrt {n^{2}+1}}-n|\leq {\frac {1}{2n}}}$

But now, ${\displaystyle \left({\tfrac {1}{2n}}\right)_{n\in \mathbb {N} }}$ is a null sequence, so the squeeze theorem proves our assertion.

## Squeeze theorem: examples and problems

#### Squeeze theorem: example & exercise 1

Example (Squeeze theorem 1)

Let us consider the sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ with

${\displaystyle a_{n}={\frac {(-1)^{n}}{n^{2}}}}$
The first 6 elements of the sequence ${\displaystyle {\color {Blue}\left({\tfrac {(-1)^{n}}{n^{2}}}\right)_{n\in \mathbb {N} }}}$, ${\displaystyle {\color {red}\left(-{\tfrac {1}{n}}\right)_{n\in \mathbb {N} }}}$ and ${\displaystyle {\color {green}\left({\tfrac {1}{n}}\right)_{n\in \mathbb {N} }}}$

We take a look at the first sequence elements ${\displaystyle (-1,{\tfrac {1}{4}},-{\tfrac {1}{9}},{\tfrac {1}{16}},-{\tfrac {1}{25}},\ldots )}$ . They seem to be alternately positive and negative and they tend to zero. We ise as lower or upper bound ${\displaystyle b_{n}=-{\tfrac {1}{n}}}$ and ${\displaystyle c_{n}={\tfrac {1}{n}}}$. Then, ${\displaystyle |a_{n}|={\tfrac {1}{n^{2}}}\leq {\tfrac {1}{n}}}$ dso we caught ${\displaystyle b_{n}\leq a_{n}\leq c_{n}}$ for all ${\displaystyle n\in \mathbb {N} }$.

In addition ${\displaystyle \lim _{n\to \infty }b_{n}=\lim _{n\to \infty }c_{n}=0}$. The squeeze theorem hence implies ${\displaystyle \lim _{n\to \infty }a_{n}=\lim _{n\to \infty }{\tfrac {(-1)^{n}}{n^{2}}}=0}$.

Question: Which further sequences ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (c_{n})_{n\in \mathbb {N} }}$ can be used for catching and squeezing ${\displaystyle a_{n}}$?

We could also use ${\displaystyle b_{n}=-{\tfrac {1}{n^{2}}}}$ and ${\displaystyle c_{n}={\tfrac {1}{n^{2}}}}$. Or any other power between 0 and 2 (not necessarily 1 or 2).

Exercise (Squeeze theorem 1)

Investigate whether the sequence ${\displaystyle ({\tilde {a}}_{n})_{n\in \mathbb {N} }}$ converges, using the squeeze theorem, where

${\displaystyle {\tilde {a}}_{n}={\frac {1}{n^{s}}}}$ with ${\displaystyle s\in \mathbb {Q} ^{+}}$

Solution (Squeeze theorem 1)

For ${\displaystyle s\in \mathbb {Q} ^{+}}$ there is a ${\displaystyle k\in \mathbb {N} }$ with ${\displaystyle k-1\leq s\leq k}$. By monotony of the power function,

${\displaystyle n^{k-1}\leq n^{s}\leq n^{k}}$

Hence

${\displaystyle \underbrace {\frac {1}{n^{k}}} _{={\tilde {b}}_{n}}\leq \underbrace {\frac {1}{n^{s}}} _{={\tilde {a}}_{n}}\leq \underbrace {\frac {1}{n^{k-1}}} _{={\tilde {c}}_{n}}}$

The limit theorems yield

${\displaystyle \lim _{n\to \infty }{\tilde {b}}_{n}=\lim _{n\to \infty }{\frac {1}{n^{k}}}=\lim _{n\to \infty }\left({\frac {1}{n}}\right)^{k}=0^{k}=0}$

and analogously, ${\displaystyle \lim _{n\to \infty }{\tilde {c}}_{n}=0}$. Hence, by means of the squeeze theorem ${\displaystyle \lim _{n\to \infty }{\tilde {a}}_{n}=0}$.

#### Squeeze theorem: example & exercise 2

Example (Squeeze theorem 2)

Next, we prove that the sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ with

${\displaystyle a_{n}={\frac {n!}{n^{n}}}}$

converges to 0.

The first elements of ${\displaystyle {\color {Blue}\left({\tfrac {n!}{n^{n}}}\right)_{n\in \mathbb {N} }}}$, ${\displaystyle {\color {red}\left(0\right)_{n\in \mathbb {N} }}}$ and ${\displaystyle {\color {green}\left({\tfrac {1}{n}}\right)_{n\in \mathbb {N} }}}$

This sequence is bounded from below as all elements are positive ${\displaystyle a_{n}\geq 0=b_{n}}$ for all ${\displaystyle n\in \mathbb {N} }$. An upper bound can be found by multiplying out ${\displaystyle n!}$ and ${\displaystyle n^{n}}$:

${\displaystyle a_{n}={\frac {n!}{n^{n}}}={\frac {\overbrace {1\cdot 2\cdot 3\cdot \ldots \cdot n} ^{n{\text{ factors}}}}{\underbrace {n\cdot n\cdot n\cdot \ldots \cdot n} _{n{\text{ factors}}}}}={\frac {1}{n}}\cdot \underbrace {\frac {2\cdot 3\cdot \ldots \cdot n}{n\cdot n\cdot \ldots \cdot n}} _{\leq 1{\text{ for }}n\geq 2}\leq \underbrace {\frac {1}{n}} _{=c_{n}}{\text{ für }}n\geq 2}$

Obviously, both bounding sequences converge to zero: ${\displaystyle \lim _{n\to \infty }b_{n}=\lim _{n\to \infty }c_{n}=0}$ . So the squeeze theorem implies ${\displaystyle \lim _{n\to \infty }a_{n}=\lim _{n\to \infty }{\tfrac {n!}{n^{n}}}=0}$.

Exercise (Squeeze theorem 2)

Use the squeeze theorem to show that ${\displaystyle ({\tilde {a}}_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle ({\hat {a}}_{n})_{n\in \mathbb {N} }}$ with

1. ${\displaystyle {\tilde {a}}_{n}={\frac {3^{n}}{n!}}}$
2. ${\displaystyle {\hat {a}}_{n}={\frac {2n}{2^{n}}}}$

both converge to 0.

Solution (Squeeze theorem 2)

Part 1: The trick is multiplying out again. this time, for ${\displaystyle 3^{n}}$ and ${\displaystyle n!}$ :

${\displaystyle \underbrace {0} _{={\tilde {b}}_{n}}\leq {\tilde {a}}_{n}={\frac {3^{n}}{n!}}={\frac {\overbrace {3\cdot 3\cdot 3\cdot \ldots \cdot 3} ^{n{\text{ factors}}}}{\underbrace {1\cdot 2\cdot 3\cdot \ldots \cdot n} _{n{\text{ factors}}}}}={\frac {3\cdot 3}{1\cdot 2}}\cdot \underbrace {\frac {3\cdot 3\cdot \ldots \cdot 3}{3\cdot 4\cdot \ldots \cdot n-1}} _{\leq 1{\text{ for }}n\geq 3}\cdot {\frac {3}{n}}\leq {\frac {9}{2}}\cdot {\frac {3}{n}}=\underbrace {\frac {27}{2n}} _{={\tilde {c}}_{n}}{\text{ for }}n\geq 3}$

Since ${\displaystyle \lim _{n\to \infty }{\tilde {b}}_{n}=\lim _{n\to \infty }{\tilde {c}}_{n}=0}$ , the squeeze theorem implies ${\displaystyle \lim _{n\to \infty }{\tilde {a}}_{n}=\lim _{n\to \infty }{\tfrac {3^{n}}{n!}}=0}$.

Part 2:

Again, the elements are positive, i.e. ${\displaystyle {\hat {a}}_{n}={\tfrac {2n}{2^{n}}}\geq 0={\hat {b}}_{n}}$. Bounding from above can be done using ${\displaystyle 2^{n}\geq n^{2}}$ for all ${\displaystyle n\geq 4}$. This inequality can be shown by induction:

Induction basis: ${\displaystyle n=4}$.

${\displaystyle 2^{4}=16\geq 16=4^{2}}$

Induction step: ${\displaystyle n\to n+1}$.

${\displaystyle 2^{n+1}=2\cdot 2^{n}{\overset {\text{IV}}{\geq }}2\cdot n^{2}=n^{2}+n^{2}=n^{2}+n\cdot n=n^{2}+2n+\underbrace {(n-2)} _{\geq 2}\cdot \underbrace {n} _{\geq 4}\geq n^{2}+2n+8\geq n^{2}+2n+1=(n+1)^{2}}$

So for all ${\displaystyle n\geq 4}$,

${\displaystyle {\hat {a}}_{n}={\frac {2n}{2^{n}}}{\overset {2^{n}\geq n^{2}}{\leq }}{\frac {2n}{n^{2}}}=\underbrace {\frac {2}{n}} _{={\hat {c}}_{n}}}$

Since ${\displaystyle \lim _{n\to \infty }{\hat {c}}_{n}=0}$ , the squeeze theorem implies ${\displaystyle \lim _{n\to \infty }{\hat {a}}_{n}=0}$.

#### Squeeze theorem: example & exercise 3

Example (Squeeze theorem 3)

Now, let us consider the root sequences ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (a_{n}')_{n\in \mathbb {N} }}$ with

${\displaystyle a_{n}={\sqrt[{n}]{n^{2}+n}}}$

and

${\displaystyle a_{n}'={\sqrt[{n}]{2^{n}+3^{n}}}}$

First sequence: For ${\displaystyle (a_{n})}$ , monotony of the ${\displaystyle n}$-th root implies

${\displaystyle a_{n}={\sqrt[{n}]{n^{2}+n}}\ {\overset {n\leq n^{2}}{\leq }}\ {\sqrt[{n}]{n^{2}+n^{2}}}=\underbrace {\sqrt[{n}]{2n^{2}}} _{=c_{n}}}$

as well as

${\displaystyle a_{n}={\sqrt[{n}]{n^{2}+n}}\ {\overset {n\geq 0}{\geq }}\ \underbrace {\sqrt[{n}]{n^{2}}} _{=b_{n}}}$

For those two bounding sequences, the limit theorems imply

${\displaystyle \lim _{n\to \infty }b_{n}=\lim _{n\to \infty }{\sqrt[{n}]{n^{2}}}=\lim _{n\to \infty }{\sqrt[{n}]{n}}\cdot {\sqrt[{n}]{n}}=1\cdot 1=1}$

and

${\displaystyle \lim _{n\to \infty }c_{n}=\lim _{n\to \infty }{\sqrt[{n}]{2n^{2}}}=\lim _{n\to \infty }{\sqrt[{n}]{2}}\cdot {\sqrt[{n}]{n}}\cdot {\sqrt[{n}]{n}}=1\cdot 1\cdot 1=1}$

The squeeze theorem hence implies ${\displaystyle \lim _{n\to \infty }a_{n}=1}$.

Second sequence: For ${\displaystyle (a_{n}')}$ the root function is monotonous, as well

${\displaystyle a_{n}'={\sqrt[{n}]{2^{n}+3^{n}}}\ {\overset {2^{n}\leq 3^{n}}{\leq }}\ {\sqrt[{n}]{3^{n}+3^{n}}}=\underbrace {\sqrt[{n}]{2\cdot 3^{n}}} _{=c_{n}'}}$

and

${\displaystyle a_{n}'={\sqrt[{n}]{2^{n}+3^{n}}}\ {\overset {2^{n}\geq 0}{\geq }}\ \underbrace {\sqrt[{n}]{3^{n}}} _{=b_{n}'}}$

We use the limit theorems for the upper and lower bounding sequences

${\displaystyle \lim _{n\to \infty }b_{n}'=\lim _{n\to \infty }{\sqrt[{n}]{3^{n}}}=\lim _{n\to \infty }3=3}$

and

${\displaystyle \lim _{n\to \infty }c_{n}'=\lim _{n\to \infty }{\sqrt[{n}]{2\cdot 3^{n}}}=\lim _{n\to \infty }{\sqrt[{n}]{2}}\cdot {\sqrt[{n}]{3^{n}}}=1\cdot 3=3}$

So the squeeze theorem implies ${\displaystyle \lim _{n\to \infty }a_{n}'=3}$.

Exercise (Squeeze theorem 3)

Use the squeeze theorem to determine the limits of the sequences ${\displaystyle ({\tilde {a}}_{n})_{n\in \mathbb {N} }}$, ${\displaystyle ({\hat {a}}_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle ({\overline {a}}_{n})_{n\in \mathbb {N} }}$ with

1. ${\displaystyle {\hat {a}}_{n}={\sqrt[{n}]{n^{2}-n+1}}}$
2. ${\displaystyle {\overline {a}}_{n}={\sqrt[{n}]{3^{n}-2^{n}}}}$

Solution (Squeeze theorem 3)

Part 1: For ${\displaystyle ({\hat {a}}_{n})}$ the ${\displaystyle n}$-th root is monotonous, so

${\displaystyle \underbrace {\sqrt[{n}]{n^{2}-n+1}} _{={\hat {a}}_{n}}\ {\overset {-n+1\leq 0}{\leq }}\ \underbrace {\sqrt[{n}]{n^{2}}} _{={\hat {c}}_{n}}}$

further, for ${\displaystyle n\geq 2}$

${\displaystyle n^{2}\geq 2n\iff {\frac {n^{2}}{2}}\geq n\iff -{\frac {n^{2}}{2}}\leq -n\iff -n+1\geq -{\frac {n^{2}}{2}}}$

Hence, there is

${\displaystyle \underbrace {\sqrt[{n}]{n^{2}-n+1}} _{={\hat {a}}_{n}}\ {\overset {-n+1\geq -{\frac {n^{2}}{2}}}{\geq }}\ {\sqrt[{n}]{n^{2}-{\frac {n^{2}}{2}}}}=\underbrace {\sqrt[{n}]{\frac {n^{2}}{2}}} _{={\hat {b}}_{n}}}$

The limit theorems for the bounding sequences yield

${\displaystyle \lim _{n\to \infty }{\hat {b}}_{n}=\lim _{n\to \infty }{\sqrt[{n}]{\frac {n^{2}}{2}}}=\lim _{n\to \infty }{\sqrt[{n}]{\frac {1}{2}}}\cdot {\sqrt[{n}]{n}}^{2}=1\cdot 1^{2}=1}$

and

${\displaystyle \lim _{n\to \infty }{\hat {c}}_{n}=\lim _{n\to \infty }{\sqrt[{n}]{n^{2}}}=\lim _{n\to \infty }{\sqrt[{n}]{n}}^{2}=1^{2}=1}$

So the squeeze theorem implies ${\displaystyle \lim _{n\to \infty }{\hat {a}}_{n}=1}$.

Part 2: For ${\displaystyle ({\overline {a}}_{n})}$ we again have monotony of the ${\displaystyle n}$-th root

${\displaystyle \underbrace {\sqrt[{n}]{3^{n}-2^{n}}} _{={\overline {a}}_{n}}\ {\overset {-2^{n}\leq 0}{\leq }}\ {\sqrt[{n}]{3^{n}}}=\underbrace {3} _{={\overline {c}}_{n}}}$

Further,

${\displaystyle \underbrace {\sqrt[{n}]{3^{n}-2^{n}}} _{={\overline {a}}_{n}}={\sqrt[{n}]{3^{n}(1-\left({\tfrac {2}{3}}\right)^{n})}}\ {\overset {-\left({\tfrac {2}{3}}\right)^{n}\geq -{\tfrac {2}{3}}}{\geq }}\ {\sqrt[{n}]{3^{n}\cdot (1-{\tfrac {2}{3}})}}=\underbrace {\sqrt[{n}]{3^{n}\cdot {\tfrac {1}{3}}}} _{={\overline {b}}_{n}}}$

The limit theorems for the bounding sequences render

${\displaystyle \lim _{n\to \infty }{\overline {b}}_{n}=\lim _{n\to \infty }{\sqrt[{n}]{3^{n}\cdot {\tfrac {1}{3}}}}=\lim _{n\to \infty }{\sqrt[{n}]{\frac {1}{3}}}\cdot {\sqrt[{n}]{3^{n}}}=1\cdot 3=3}$

And by means of the squeeze theorem ${\displaystyle \lim _{n\to \infty }{\overline {a}}_{n}=3}$.

#### Squeeze theorem: example & exercise 4

Example (Squeeze theorem 4)

Let ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ be the sequence of partial sums

${\displaystyle a_{n}=\sum _{k=1}^{n}{\frac {k}{n^{2}+k}}}$

Then, for all ${\displaystyle k=1,\ldots ,n}$ we bound:

${\displaystyle {\frac {k}{n^{2}+n}}{\overset {k\leq n}{\leq }}{\frac {k}{n^{2}+k}}{\overset {k\geq 0}{\leq }}{\frac {k}{n^{2}}}}$

and hence

${\displaystyle \underbrace {\sum _{k=1}^{n}{\frac {k}{n^{2}+n}}} _{=b_{n}}{\overset {k\leq n}{\leq }}\underbrace {\sum _{k=1}^{n}{\frac {k}{n^{2}+k}}} _{=a_{n}}{\overset {k\geq 0}{\leq }}\underbrace {\sum _{k=1}^{n}{\frac {k}{n^{2}}}} _{=c_{n}}}$

For the lower bound,

{\displaystyle {\begin{aligned}b_{n}&=\sum _{k=1}^{n}{\frac {k}{n^{2}+n}}\\[0.5em]&={\frac {\sum _{k=1}^{n}k}{n^{2}+n}}\\[0.5em]&\ {\color {OliveGreen}\left\downarrow \ {\text{Gaussian summation }}\sum _{k=1}^{n}k={\frac {n(n+1)}{2}}\right.}\\[0.5em]&={\frac {\frac {n(n+1)}{2}}{n^{2}+n}}\\[0.5em]&={\frac {n(n+1)}{2(n^{2}+n)}}\\[0.5em]&={\frac {n^{2}+n}{2n^{2}+2n}}\\[0.5em]&={\frac {1+{\frac {1}{n}}}{2+{\frac {2}{n}}}}\\[0.5em]&\to {\frac {1}{2}}{\text{ with }}n\to \infty \end{aligned}}}

For the upper bound,

{\displaystyle {\begin{aligned}c_{n}&=\sum _{k=1}^{n}{\frac {k}{n^{2}}}\\[0.5em]&={\frac {\sum _{k=1}^{n}k}{n^{2}}}\\[0.5em]&\ {\color {OliveGreen}\left\downarrow \ {\text{Gaussian summation }}\sum _{k=1}^{n}k={\frac {n(n+1)}{2}}\right.}\\[0.5em]&={\frac {\frac {n(n+1)}{2}}{n^{2}}}\\[0.5em]&={\frac {n(n+1)}{2n^{2}}}\\[0.5em]&={\frac {n^{2}+n}{2n^{2}}}\\[0.5em]&={\frac {1+{\frac {1}{n}}}{2}}\\[0.5em]&\to {\frac {1}{2}}{\text{ with }}n\to \infty \end{aligned}}}

So the squeeze theorem implies ${\displaystyle \lim _{n\to \infty }a_{n}={\tfrac {1}{2}}}$.

Exercise (Squeeze theorem 4)

Investigate the limits of the two sequences of partial sums ${\displaystyle ({\tilde {a}}_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle ({\hat {a}}_{n})_{n\in \mathbb {N} }}$ defined by

1. ${\displaystyle {\tilde {a}}_{n}=\sum _{k=1}^{n}{\frac {k^{2}}{n^{3}+k}}}$
2. ${\displaystyle {\hat {a}}_{n}=\sum _{k=1}^{n}{\frac {k^{2}}{n^{4}-10k^{2}}}}$

using the squeeze theorem.

Solution (Squeeze theorem 4)

Part 1: For all ${\displaystyle k=1,\ldots ,n}$ we have:

${\displaystyle {\frac {k^{2}}{n^{3}+n}}{\overset {k\leq n}{\leq }}{\frac {k^{2}}{n^{3}+k}}{\overset {k\geq 0}{\leq }}{\frac {k^{2}}{n^{3}}}}$

so we have some bounds

${\displaystyle \underbrace {\sum _{k=1}^{n}{\frac {k^{2}}{n^{3}}}} _{={\tilde {b}}_{n}}{\overset {k\geq 0}{\leq }}\underbrace {\sum _{k=1}^{n}{\frac {k^{2}}{n^{3}+k}}} _{={\tilde {a}}_{n}}{\overset {k\leq n}{\leq }}\underbrace {\sum _{k=1}^{n}{\frac {k^{2}}{n^{3}+n}}} _{={\tilde {c}}_{n}}}$

For the upper bound,

{\displaystyle {\begin{aligned}{\tilde {b}}_{n}&=\sum _{k=1}^{n}{\frac {k^{2}}{n^{3}+n}}\\[0.5em]&={\frac {\sum _{k=1}^{n}k^{2}}{n^{3}+n}}\\[0.5em]&\ {\color {OliveGreen}\left\downarrow \ {\text{sum formula }}\sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}\right.}\\[0.5em]&={\frac {\frac {n(n+1)(2n+1)}{6}}{n^{3}+n}}\\[0.5em]&={\frac {n(n+1)(2n+1)}{6(n^{3}+n)}}\\[0.5em]&={\frac {(n^{2}+n)(2n+1)}{6n^{3}+6n}}\\[0.5em]&={\frac {2n^{3}+3n^{2}+n}{6n^{3}+6n}}\\[0.5em]&={\frac {2+{\frac {3}{n}}+{\frac {1}{n^{2}}}}{6+{\frac {6}{n^{2}}}}}\\[0.5em]&\to {\frac {2}{6}}={\frac {1}{3}}{\text{ with }}n\to \infty \end{aligned}}}

For the lower bound,

{\displaystyle {\begin{aligned}{\tilde {c}}_{n}&=\sum _{k=1}^{n}{\frac {k^{2}}{n^{3}}}\\[0.5em]&={\frac {\sum _{k=1}^{n}k^{2}}{n^{3}}}\\[0.5em]&\ {\color {OliveGreen}\left\downarrow \ {\text{sum formula }}\sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}\right.}\\[0.5em]&={\frac {\frac {n(n+1)(2n+1)}{6}}{n^{3}}}\\[0.5em]&={\frac {n(n+1)(2n+1)}{6n^{3}}}\\[0.5em]&={\frac {(n^{2}+n)(2n+1)}{6n^{3}}}\\[0.5em]&={\frac {2n^{3}+3n^{2}+n}{6n^{3}}}\\[0.5em]&={\frac {2+{\frac {3}{n}}+{\frac {1}{n^{2}}}}{6}}\\[0.5em]&\to {\frac {2}{6}}={\frac {1}{3}}{\text{ with }}n\to \infty \end{aligned}}}

So the squeeze theorem implies ${\displaystyle \lim _{n\to \infty }{\tilde {a}}_{n}={\tfrac {1}{3}}}$.

Part 2: For all ${\displaystyle n\geq 4}$ and ${\displaystyle k=1,\ldots ,n}$ there is:

${\displaystyle 0\leq {\frac {k^{2}}{n^{4}-10k^{2}}}{\overset {-10k^{2}\geq -10n^{2}}{\leq }}{\frac {k^{2}}{n^{4}-10n^{2}}}}$

so we have the bounds

${\displaystyle \underbrace {0} _{={\hat {b}}_{n}}{\overset {k\geq 0}{\leq }}\underbrace {\sum _{k=1}^{n}{\frac {k^{2}}{n^{4}-10k^{2}}}} _{={\hat {a}}_{n}}\leq \underbrace {\sum _{k=1}^{n}{\frac {k^{2}}{n^{4}-10n^{2}}}} _{={\hat {c}}_{n}}}$

The lower bound is just 0 and for the upper bound,

{\displaystyle {\begin{aligned}{\hat {c}}_{n}&=\sum _{k=1}^{n}{\frac {k^{2}}{n^{4}-10n^{2}}}\\[0.5em]&={\frac {\sum _{k=1}^{n}k^{2}}{n^{4}-10n^{2}}}\\[0.5em]&\ {\color {OliveGreen}\left\downarrow \ {\text{sum formula }}\sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}\right.}\\[0.5em]&={\frac {\frac {n(n+1)(2n+1)}{6}}{n^{4}-10n^{2}}}\\[0.5em]&={\frac {2n^{3}+3n^{2}+n}{6n^{4}-60n^{2}}}\\[0.5em]&={\frac {{\frac {2}{n}}+{\frac {3}{n^{2}}}+{\frac {1}{n^{3}}}}{6-{\frac {60}{n^{2}}}}}\\[0.5em]&\to 0{\text{ with }}n\to \infty \end{aligned}}}

So the squeeze theorem implies ${\displaystyle \lim _{n\to \infty }{\hat {a}}_{n}=0}$.

#### Squeeze theorem: example & exercise 5

Exercise (Squeeze theorem 5)

Does the series ${\displaystyle a_{n}=\left(1-{\tfrac {1}{n^{2}}}\right)^{n}}$ converge`? If yes, what is its limit? Provide proofs for your claims.

How to get to the proof? (Squeeze theorem 5)

Perhaps, you have already encountered the sequence ${\displaystyle d_{n}=\left(1+{\tfrac {1}{n}}\right)^{n}}$ which converges to Euler's number ${\displaystyle e}$. This is an upper bound for ${\displaystyle a_{n}=\left(1-{\tfrac {1}{n^{2}}}\right)^{n}}$ , so with that knowledge, we may assert that it converges. But what is the limit? Maybe, we should start by computing some initial sequence elements.:

{\displaystyle {\begin{aligned}a_{10}&=0{,}90438\ldots \\a_{100}&=0{,}99004\ldots \\a_{1000}&=0{,}99900\ldots \end{aligned}}}

It looks as if ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ converges to ${\displaystyle 1}$ . This gets particularly clear, if we plot the sequence elements in a diagram:

What makes determining the limit hard is the exponent of ${\displaystyle n}$ . Otherwise, we could use some limit theorems of an epsilon-delta proof and would be done quite fast. Luckily, there is a trick to get rid of the exponent: Bernoulli's inequality:

${\displaystyle \left(1-{\frac {1}{n^{2}}}\right)^{n}\geq 1-n\cdot {\frac {1}{n^{2}}}=1-{\frac {1}{n}}}$

The sequence ${\displaystyle \left(1-{\tfrac {1}{n}}\right)_{n\in \mathbb {N} }}$ is hence a lower bound on ${\displaystyle a_{n}}$ and it clearly converges to ${\displaystyle 1}$ . We hence set ${\displaystyle b_{n}=1-{\tfrac {1}{n}}}$ as a lower bound for the squeeze theorem.

How to get an upper bound converging to 1? This is quite easy, since ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is always smaller than ${\displaystyle 1}$ : The expression ${\displaystyle 1-{\tfrac {1}{n^{2}}}}$ is always smaller than 1 and therefore, also every power of it will be. This includes ${\displaystyle \left(1-{\tfrac {1}{n^{2}}}\right)^{n}}$ being smaller than ${\displaystyle 1}$.

So for the squeeze theorem, we have for all ${\displaystyle n\in \mathbb {N} }$ the inequality:

${\displaystyle 1-{\frac {1}{n}}\leq \left(1-{\frac {1}{n^{2}}}\right)^{n}\leq 1}$

both the lower and the upper bounding sequence converge to the identical limit ${\displaystyle 1}$ . So by means of the squeeze theorem, there must also be ${\displaystyle \lim _{n\rightarrow \infty }\left(1-{\tfrac {1}{n^{2}}}\right)^{n}=1}$.

Proof (Squeeze theorem 5)

Bernoulli's inequality implies

${\displaystyle \left(1-{\frac {1}{n^{2}}}\right)^{n}\geq 1-n\cdot {\frac {1}{n^{2}}}=1-{\frac {1}{n}}}$

And there is

${\displaystyle {\frac {1}{n^{2}}}\geq 0\Rightarrow 1-{\frac {1}{n^{2}}}\leq 1\Rightarrow \left(1-{\frac {1}{n^{2}}}\right)^{n}\leq 1}$

So we get the bounds

${\displaystyle 1-{\frac {1}{n}}\leq \left(1-{\frac {1}{n^{2}}}\right)^{n}\leq 1}$

Both bounding sequences converge to ${\displaystyle \lim _{n\rightarrow \infty }1-{\tfrac {1}{n}}=\lim _{n\rightarrow \infty }1=1}$ , so by means of the squeeze theorem, ${\displaystyle \lim _{n\rightarrow \infty }\left(1-{\tfrac {1}{n^{2}}}\right)^{n}=1}$ which was to be shown.

Hint

If you are allowed to use ${\displaystyle \lim _{n\to \infty }\left(1+{\tfrac {1}{n}}\right)^{n}=e}$ and ${\displaystyle \lim _{n\to \infty }\left(1-{\tfrac {1}{n}}\right)^{n}={\tfrac {1}{e}}}$ , then you can also obtain the result directly, using the third binomial formula:

{\displaystyle {\begin{aligned}\lim _{n\to \infty }\left(1-{\tfrac {1}{n^{2}}}\right)^{n}&=\lim _{n\to \infty }\left(\left(1+{\tfrac {1}{n}}\right)\left(1-{\tfrac {1}{n}}\right)\right)^{n}\\&=\lim _{n\to \infty }\left(1+{\tfrac {1}{n}}\right)^{n}\left(1-{\tfrac {1}{n}}\right)^{n}\\&=\lim _{n\to \infty }\left(1+{\tfrac {1}{n}}\right)^{n}\cdot \lim _{n\to \infty }\left(1-{\tfrac {1}{n}}\right)^{n}\\&=e\cdot {\tfrac {1}{e}}=1\end{aligned}}}

Exercise (Squeeze theorem 5)

Does the sequence ${\displaystyle a_{n}=\left(1+{\tfrac {1}{n^{2}}}\right)^{n}}$ converge? If yes, what is its limit? Prove your assertions.

How to get to the proof? (Squeeze theorem 5)

Again, Bernoulli's inequality can be used for getting rid of the exponential ${\displaystyle n}$. There is

${\displaystyle \left(1+{\frac {1}{n^{2}}}\right)^{n}\geq 1+n\cdot {\frac {1}{n^{2}}}=1+{\frac {1}{n}}}$

so we have a bound from below converging to ${\displaystyle 1}$. But ${\displaystyle \left(1+{\tfrac {1}{n^{2}}}\right)^{n}\geq 1}$ is already a lower bound converging to ${\displaystyle 1}$. So Bernoulli's inequality doesn't help us in this form. What we need is an upper bound converging to ${\displaystyle 1}$. This can be done by a little trick: we put expressions from the enumerator in the denominator and apply Bernoulli's inequality afterwards:

{\displaystyle {\begin{aligned}\left(1+{\frac {1}{n^{2}}}\right)^{n}&=\left({\frac {n^{2}+1}{n^{2}}}\right)^{n}\\[0.3em]&={\frac {1}{\left({\frac {n^{2}}{n^{2}+1}}\right)^{n}}}\\[0.3em]&={\frac {1}{\left({\frac {n^{2}+1-1}{n^{2}+1}}\right)^{n}}}\\[0.3em]&={\frac {1}{\left(1-{\frac {1}{n^{2}+1}}\right)^{n}}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{Bernoulli's inequality}}\right.}\\[0.3em]&\leq {\frac {1}{\left(1-{\frac {n}{n^{2}+1}}\right)}}\\[0.3em]&={\frac {1}{\left({\tfrac {n^{2}-n+1}{n^{2}+1}}\right)}}\\[0.3em]&={\frac {n^{2}+1}{n^{2}-n+1}}\end{aligned}}}

Hence, we have an upper bound ${\displaystyle a_{n}\leq {\tfrac {n^{2}+1}{n^{2}-n+1}}}$. Enumerator and denominator of the bound are both dominated by the ${\displaystyle n^{2}}$, so they cancel out when using the limit theorems and we obtain ${\displaystyle \lim _{n\to \infty }{\tfrac {n^{2}+1}{n^{2}-n+1}}=1}$. Therefore, we have two bounding sequences ${\displaystyle b_{n}=1\leq \left(1+{\tfrac {1}{n^{2}}}\right)^{n}\leq {\tfrac {n^{2}+1}{n^{2}-n+1}}=c_{n}}$ converging to ${\displaystyle 1}$ and by the squeeze theorem ${\displaystyle \lim _{n\to \infty }a_{n}=1}$ .

Proof (Squeeze theorem 5)

Let ${\displaystyle b_{n}=1}$ and ${\displaystyle c_{n}={\tfrac {n^{2}+1}{n^{2}-n+1}}}$. There is ${\displaystyle b_{n}\leq a_{n}\leq c_{n}}$:

{\displaystyle {\begin{aligned}1&\leq \left(1+{\frac {1}{n^{2}}}\right)^{n}\\[0.3em]&=\left({\frac {n^{2}+1}{n^{2}}}\right)^{n}\\[0.3em]&={\frac {1}{\left({\frac {n^{2}}{n^{2}+1}}\right)^{n}}}\\[0.3em]&={\frac {1}{\left({\frac {n^{2}+1-1}{n^{2}+1}}\right)^{n}}}\\[0.3em]&={\frac {1}{\left(1-{\frac {1}{n^{2}+1}}\right)^{n}}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{Bernoulli's inequality}}\right.}\\[0.3em]&\leq {\frac {1}{\left(1-{\frac {n}{n^{2}+1}}\right)}}\\[0.3em]&={\frac {1}{\left({\tfrac {n^{2}-n+1}{n^{2}+1}}\right)}}\\[0.3em]&={\frac {n^{2}+1}{n^{2}-n+1}}\end{aligned}}}

The bounding sequences converge to ${\displaystyle \lim _{n\to \infty }b_{n}=\lim _{n\to \infty }1=1}$ and (by means of the limit theorem):

{\displaystyle {\begin{aligned}\lim _{n\to \infty }c_{n}&=\lim _{n\to \infty }{\frac {n^{2}+1}{n^{2}-n+1}}\\[0.3em]&=\lim _{n\to \infty }{\frac {1+{\tfrac {1}{n^{2}}}}{1+{\frac {1}{n}}+{\frac {1}{n^{2}}}}}\\[0.3em]&={\frac {1+0}{1-0+0}}=1\end{aligned}}}

So there is ${\displaystyle \lim _{n\to \infty }b_{n}=\lim _{n\to \infty }c_{n}=1}$ and the squeeze theorem implies ${\displaystyle \lim _{n\to \infty }a_{n}=1}$ .

Alternative proof (Squeeze theorem 5)

If the exponential series ${\displaystyle \sum _{k=0}^{\infty }{\tfrac {1}{k!}}=e}$ is known, one can also use the binomial theorem for a bounding. A lower bound is given by ${\displaystyle \left(1+{\tfrac {1}{n^{2}}}\right)^{n}\geq 1}$ . For the upper bound, we use the binomial theorem ${\displaystyle (a+b)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}a^{k}b^{n-k}}$:

{\displaystyle {\begin{aligned}\left(1+{\frac {1}{n^{2}}}\right)^{n}&=\sum _{k=0}^{n}{\binom {n}{k}}\left({\frac {1}{n^{2}}}\right)^{k}\cdot 1^{n-k}\\[0.3em]&=1+\sum _{k=1}^{n}{\frac {n\cdot (n-1)\cdot \ldots \cdot (n-k+1)}{k!}}\cdot {\frac {1}{n^{2k}}}\\[0.3em]&=1+\sum _{k=1}^{n}{\frac {1}{k!}}\cdot \underbrace {\frac {n\cdot (n-1)\cdot \ldots \cdot (n-k+1)}{n\cdot n\cdot \ldots \cdot n}} _{\leq 1}\cdot \underbrace {\frac {1}{n^{k}}} _{\leq {\frac {1}{n}}}\\[0.3em]&\leq 1+{\frac {1}{n}}\cdot \sum _{k=1}^{n}{\frac {1}{k!}}\\[0.3em]&\leq 1+{\frac {1}{n}}\cdot \underbrace {\sum _{k=1}^{\infty }{\frac {1}{k!}}} _{=e-1}\\[0.3em]&\leq 1+{\frac {e-1}{n}}\end{aligned}}}

This yields a suitable upper bound and we get ${\displaystyle 1\leq a_{n}\leq 1+{\tfrac {e-1}{n}}}$ . The squeeze theorem then implies that ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ converges to ${\displaystyle 1}$ .

Alternative proof (Squeeze theorem 5)

We may use ${\displaystyle \lim _{n\to \infty }\left(1+{\tfrac {1}{n}}\right)^{n}=e}$ for a direct proof. The sequence ${\displaystyle \left(\left(1+{\tfrac {1}{n^{2}}}\right)^{n^{2}}\right)_{n\in \mathbb {N} }}$ is a subsequence of ${\displaystyle \left(\left(1+{\tfrac {1}{n}}\right)^{n}\right)_{n\in \mathbb {N} }}$ and converges to the same limit:

${\displaystyle \lim _{n\to \infty }\left(1+{\tfrac {1}{n^{2}}}\right)^{n^{2}}=e}$

Hence, for some ${\displaystyle N\in \mathbb {N} }$, there is ${\displaystyle \left(1+{\tfrac {1}{n^{2}}}\right)^{n^{2}}\leq 2e}$ for all ${\displaystyle n\geq N}$. The sequence ${\displaystyle a_{n}}$ can be recovered from this by taking the ${\displaystyle n^{th}}$ root ${\displaystyle \left(1+{\tfrac {1}{n^{2}}}\right)^{n}={\sqrt[{n}]{\left(1+{\tfrac {1}{n^{2}}}\right)^{n^{2}}}}}$ . So for all ${\displaystyle n\geq N}$:

{\displaystyle {\begin{aligned}{\begin{array}{rrcl}&1\leq &\left(1+{\tfrac {1}{n^{2}}}\right)^{n^{2}}&\leq 2e\\[0.5em]\implies &{\sqrt[{n}]{1}}\leq &{\sqrt[{n}]{\left(1+{\tfrac {1}{n^{2}}}\right)^{n^{2}}}}&\leq {\sqrt[{n}]{2e}}\\[0.5em]\implies &1\leq &\left(1+{\tfrac {1}{n^{2}}}\right)^{n}&\leq {\sqrt[{n}]{2e}}\end{array}}\end{aligned}}}

With ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{1}}=1=\lim _{n\to \infty }{\sqrt[{n}]{2e}}}$ , the squeeze theorem again implies that ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ converges to ${\displaystyle 1}$.

## Examples & exercises: squeeze theorem for null sequences

Example (Squeeze theorem for null sequences 1)

First, an easy example: we prove ${\displaystyle \lim _{n\to \infty }{\tfrac {n-1}{n+1}}=1}$.

We estimate ${\displaystyle |\underbrace {\tfrac {n-1}{n+1}} _{=a_{n}}-\underbrace {1} _{=a}|}$ by the null sequence ${\displaystyle c_{n}={\tfrac {2}{n}}}$ as follows:

${\displaystyle \left|{\frac {n-1}{n+1}}-1\right|=\left|{\frac {n-1-(n+1)}{n+1}}\right|=\left|{\frac {-2}{n+1}}\right|={\frac {2}{n+1}}\leq {\frac {2}{n}}}$

Since ${\displaystyle \lim _{n\to \infty }c_{n}=\lim _{n\to \infty }{\tfrac {2}{n}}=0}$ is a null sequence, we can use ${\displaystyle c_{n}}$ as an upper and ${\displaystyle -c_{n}}$ as a lower bound for ${\displaystyle a_{n}}$. The squeeze theorem then implies ${\displaystyle \lim _{n\to \infty }a_{n}=\lim _{n\to \infty }{\tfrac {n-1}{n+1}}=1}$.

Example (Squeeze theorem for null sequences 2)

Now, it gets a bit more complicated: We fix any ${\displaystyle c>1}$ and prove that ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{c}}=1}$. This can be done by bounding ${\displaystyle |{\sqrt[{n}]{c}}-1|={\sqrt[{n}]{c}}-1}$ from above using a null sequence ${\displaystyle (c_{n})}$ . the trick is to use Bernoulli's inequality, which will yield us to the upper bound ${\displaystyle c_{n}={\tfrac {c-1}{n}}}$ :

{\displaystyle {\begin{aligned}c&=({\sqrt[{n}]{c}})^{n}\\[0.5em]&=(1+({\sqrt[{n}]{c}}-1))^{n}\\[0.5em]&{\color {OliveGreen}\left\downarrow {\text{ Bernoulli's inequality}}\right.}\\[0.5em]&\geq 1+n\cdot ({\sqrt[{n}]{c}}-1)\end{aligned}}}

Which implies

{\displaystyle {\begin{aligned}c\geq 1+n\cdot ({\sqrt[{n}]{c}}-1)&\Leftrightarrow c-1\geq n\cdot ({\sqrt[{n}]{c}}-1)\\[0.5em]&\Leftrightarrow {\tfrac {c-1}{n}}\geq {\sqrt[{n}]{c}}-1\\[0.5em]&\Leftrightarrow {\sqrt[{n}]{c}}-1\leq {\tfrac {c-1}{n}}\end{aligned}}}

So we have

${\displaystyle |{\sqrt[{n}]{c}}-1|={\sqrt[{n}]{c}}-1\leq {\tfrac {c-1}{n}}=(c-1)\cdot {\frac {1}{n}}{\xrightarrow {n\to \infty }}0}$

The squeeze theorem for null sequences directly implies ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{c}}=1}$.

Exercise (Squeeze theorem for null sequences)

Prove that ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{n}}=1}$.

Solution (Squeeze theorem for null sequences)

In order to use the squeeze theorem, we need to bound ${\displaystyle |{\sqrt[{n}]{n}}-1|}$ from above by a null sequence. The binomial theorem can be used for this bounding. In case ${\displaystyle n\geq 2}$:

{\displaystyle {\begin{aligned}n&=({\sqrt[{n}]{n}})^{n}\\[0.5em]&=(1+({\sqrt[{n}]{n}}-1))^{n}\\[0.5em]&{\color {OliveGreen}\left\downarrow {\text{ binomial theorem}}\right.}\\[0.5em]&=\sum _{k=0}^{n}{\binom {n}{k}}({\sqrt[{n}]{n}}-1)^{k}\cdot \underbrace {1^{n-k}} _{=1}\\&=\underbrace {{\binom {n}{0}}({\sqrt[{n}]{n}}-1)^{0}} _{=1}+\underbrace {{\binom {n}{1}}({\sqrt[{n}]{n}}-1)^{1}} _{\geq 0}+{\binom {n}{2}}({\sqrt[{n}]{n}}-1)^{2}+\underbrace {\sum _{k=3}^{n}{\binom {n}{k}}({\sqrt[{n}]{n}}-1)^{k}} _{\geq 0}\\[0.5em]&\geq 1+{\binom {n}{2}}({\sqrt[{n}]{n}}-1)^{2}\\[0.5em]&=1+{\frac {n(n-1)}{2}}({\sqrt[{n}]{n}}-1)^{2}\end{aligned}}}

This inequality implies

{\displaystyle {\begin{aligned}n\geq 1+{\frac {n(n-1)}{2}}\cdot ({\sqrt[{n}]{n}}-1)^{2}&\iff n-1\geq {\frac {n(n-1)}{2}}\cdot ({\sqrt[{n}]{n}}-1)^{2}\\[0.5em]&\iff {\tfrac {2}{n}}\geq ({\sqrt[{n}]{n}}-1)^{2}\\[0.5em]&\iff \left|{\sqrt[{n}]{n}}-1\right|\leq {\sqrt {\tfrac {2}{n}}}\end{aligned}}}

${\displaystyle |{\sqrt[{n}]{n}}-1|\leq {\sqrt {2}}\cdot {\frac {1}{\sqrt {n}}}{\overset {n\to \infty }{\to }}0}$
And by means of the squeeze theorem ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{n}}=1}$, which was to be shown.