# Exercises: Subsequences, Accumulation points and Cauchy sequences – Serlo

Zur Navigation springen Zur Suche springen

## Exercises: Subsequences

Exercise (Subsequences of the e-sequence)

1. Calculate the limit ${\displaystyle \lim _{n\to \infty }\left(1+{\frac {1}{n^{2}}}\right)^{n^{2}}}$
2. Calculate the limit ${\displaystyle \lim _{n\to \infty }\left(1+{\frac {1}{2n}}\right)^{n}}$
3. Calculate the limit ${\displaystyle \lim _{n\to \infty }\left(1+{\frac {1}{n^{2}}}\right)^{n}}$ using 1)

Solution (Subsequences of the e-sequence)

Subtask 1: As we know that the sequence ${\displaystyle (a_{n})=\left(\left(1+{\frac {1}{n}}\right)^{n}\right)}$ has the limit ${\displaystyle \lim _{n\to \infty }a_{n}=e}$, for the subsequence ${\displaystyle (a_{k^{2}})_{k\in \mathbb {N} }}$ holds that:

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

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

and ${\displaystyle (a_{2n})=\left(\left(1+{\frac {1}{2n}}\right)^{2n}\right)}$ being a subsequence of the sequence ${\displaystyle (a_{n})}$ from 1, we get ${\displaystyle \lim _{n\to \infty }a_{2n}=\lim _{n\to \infty }\left(1+{\frac {1}{2n}}\right)^{2n}=e}$. Using the computation rules for limits, we get

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

Subtask 3: For ${\displaystyle (b_{n})=\left(\left(1+{\frac {1}{n^{2}}}\right)^{n}\right)}$ holds that ${\displaystyle b_{n}={\sqrt[{n}]{a_{n^{2}}}}}$ with the sequence ${\displaystyle (a_{n^{2}})=\left(\left(1+{\frac {1}{n^{2}}}\right)^{n^{2}}\right)}$ from subtask 1.

As ${\displaystyle \lim _{n\to \infty }a_{n^{2}}=e}$, ${\displaystyle (a_{n^{2}})}$ is bounded from above, there exists an ${\displaystyle M>0}$ (actually ${\displaystyle M>1}$) with ${\displaystyle a_{n^{2}}\leq M}$ for all ${\displaystyle n\in \mathbb {N} }$. Therefore, ${\displaystyle {\sqrt[{n}]{M}}}$ is an upper bound of ${\displaystyle (b_{n})}$. We even get

${\displaystyle 1\leq b_{n}=\left(1+{\frac {1}{n^{2}}}\right)^{n}\leq {\sqrt[{n}]{M}}}$

As ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{M}}=1}$, using the squeeze theorem we get

${\displaystyle \lim _{k\to \infty }b_{n}=\lim _{n\to \infty }\left(1+{\frac {1}{n^{2}}}\right)^{n}=1}$

## Exercises: Accumulation points

Exercise (Accumulation points)

Determine all accumulation points of the sequences ${\displaystyle (a_{n})_{n\in \mathbb {N} },(b_{n})_{n\in \mathbb {N} }}$ with ${\displaystyle a_{n}={\tfrac {1}{2^{n}}}+(-1)^{n}}$ and ${\displaystyle b_{n}=\left(1+{\tfrac {(-1)^{n}}{n}}\right)^{n}}$.

How to get to the proof? (Accumulation points)

• Regarding ${\displaystyle (a_{n})}$: As ${\displaystyle ({\tfrac {1}{2^{n}}})}$ converges to ${\displaystyle 0}$, all of its subsequences converge to ${\displaystyle 0}$. The sequence ${\displaystyle ((-1)^{n})}$ alternates between ${\displaystyle 1}$ and ${\displaystyle -1}$. Hence, its accumulation points are ${\displaystyle 1}$ and ${\displaystyle -1}$.
• Regarding ${\displaystyle (b_{n})}$: For even ${\displaystyle n}$-s, ${\displaystyle (b_{n})}$ is a subsequence of the ${\displaystyle e}$-sequence ${\displaystyle \left(1+{\tfrac {1}{n}}\right)^{n}}$. For odd ${\displaystyle n}$-s, ${\displaystyle (b_{n})}$ is a subsequence of the sequence ${\displaystyle \left(1-{\tfrac {1}{n}}\right)^{n}}$, which converges to ${\displaystyle {\tfrac {1}{e}}}$.

Solution (Accumulation points)

• As ${\displaystyle a_{2k}={\tfrac {1}{2^{2k}}}+(-1)^{2k}={\tfrac {1}{4^{k}}}+1{\overset {k\to \infty }{\to }}0+1=1}$, one accumulation point of ${\displaystyle (a_{n})}$ is ${\displaystyle 1}$.

Furthermore, we have ${\displaystyle a_{2k-1}={\tfrac {1}{2^{2k-1}}}+(-1)^{2k-1}={\tfrac {2}{4^{k-1}}}-1{\overset {k\to \infty }{\to }}0-1=-1}$. Hence, another accumulation point of ${\displaystyle (a_{n})}$ is ${\displaystyle -1}$.

These are the only accumulation points of ${\displaystyle (a_{n})}$ because every other subsequence of ${\displaystyle (a_{n})}$ either diverges or converges to ${\displaystyle 1}$ or ${\displaystyle -1}$.

• We have ${\displaystyle b_{2k}=\left(1+{\tfrac {1}{2k}}\right)^{2k}{\overset {k\to \infty }{\to }}e}$, as ${\displaystyle \left(1+{\tfrac {1}{n}}\right)^{n}{\overset {n\to \infty }{\to }}e}$. Thus, an accumulation point of ${\displaystyle (b_{n})}$ is ${\displaystyle e}$.

Furthermore, we have ${\displaystyle b_{2k+1}=\left(1-{\tfrac {1}{2k+1}}\right)^{2k+1}{\overset {k\to \infty }{\to }}{\tfrac {1}{e}}}$, as ${\displaystyle \left(1-{\tfrac {1}{n}}\right)^{n}{\overset {n\to \infty }{\to }}{\tfrac {1}{e}}}$. Therefore, another accumulation point of ${\displaystyle (b_{n})}$ is ${\displaystyle {\tfrac {1}{e}}}$.

These are the only accumulation points of ${\displaystyle (b_{n})}$ because every other subsequence of ${\displaystyle (b_{n})}$ either diverges or converges to ${\displaystyle e}$ or ${\displaystyle {\tfrac {1}{e}}}$.

## Exercises: Lim inf and Lim sup

Exercise (Lim inf and Lim sup)

Determine Lim inf and Lim sup of the sequences ${\displaystyle (a_{n})_{n\in \mathbb {N} },(b_{n})_{n\in \mathbb {N} }}$ with ${\displaystyle a_{n}=\left({\tfrac {3}{2}}+(-1)^{n}\right)^{n}}$ and ${\displaystyle b_{n}={\begin{cases}2+{\tfrac {1}{3^{n}}}&{\text{für }}n=3k,\\3+{\tfrac {n+2}{n}}&{\text{für }}n=3k+1,\\3&{\text{für }}n=3k+2\end{cases}}}$.

How to get to the proof? (Lim inf and Lim sup)

• We start off investigating whether ${\displaystyle (a_{n})}$ is bounded from above/below. This sequence is bounded from below but not from above. Hence, Lim inf equals the smallest accumulation point of ${\displaystyle (a_{n})}$. We can determine it looking at the subsequence including the odd sequence elements. As the sequence is not bounded from above, Lim sup equals ${\displaystyle \infty }$.
• As all of its subsequences are bounded, ${\displaystyle (b_{n})}$ is bounded from above and below. In order to determine Lim sup and Lim inf of ${\displaystyle (b_{n})}$, we need to find the limits of its subsequences.

Solution (Lim inf and Lim sup)

• As ${\displaystyle {\tfrac {3}{2}}+(-1)^{n}\geq 0}$, we have ${\displaystyle a_{n}\geq 0}$. Hence, ${\displaystyle (a_{n})}$ is bounded from below. ${\displaystyle \liminf a_{n}}$ then equals the smallest accumulation point of the sequence ${\displaystyle (a_{n})}$. For the subsequence ${\displaystyle (a_{2k-1})}$ holds that
${\displaystyle \lim _{k\to \infty }a_{2k-1}=\lim _{k\to \infty }\left({\tfrac {3}{2}}+(-1)^{2k-1}\right)^{2k-1}=\lim _{k\to \infty }\left({\tfrac {3}{2}}-1\right)^{2k-1}=\lim _{k\to \infty }\left({\tfrac {1}{2}}\right)^{2k-1}=0}$

Therefore, ${\displaystyle 0}$ is an accumulation point of ${\displaystyle (a_{n})}$. There can't be other accumulation points, as all other subsequences of our sequence either converge to ${\displaystyle 0}$ or diverge. Hence, we have ${\displaystyle \liminf _{n\to \infty }a_{n}=0}$.

Alternative solution: We have ${\displaystyle ({\tilde {b}}_{k})_{k\in \mathbb {N} }=(\inf\{a_{n}:n\geq k\})_{k\in \mathbb {N} }=(0,0,0,0,\ldots )}$, as the odd sequence elements of ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ monotonically decrease to ${\displaystyle 0}$. It follows that ${\displaystyle \liminf _{n\to \infty }a_{n}=\lim _{k\to \infty }b_{k}=0}$

With ${\displaystyle n=2k}$, we get

${\displaystyle a_{2k}=\left({\tfrac {3}{2}}+(-1)^{2k}\right)^{2k}=\left({\tfrac {3}{2}}+1\right)^{2k}=\left({\tfrac {5}{2}}\right)^{2k}=\left({\tfrac {25}{4}}\right)^{k}}$

The archimedean property then states that there exists a ${\displaystyle k\in \mathbb {N} }$ with ${\displaystyle a_{2k}=\left({\tfrac {25}{4}}\right)^{k}>S}$ for all ${\displaystyle S>0}$. Therefore, ${\displaystyle (a_{n})}$ is not bounded from above and ${\displaystyle \limsup _{n\to \infty }a_{n}=\infty }$.

• Regarding ${\displaystyle (b_{n})}$: We have
{\displaystyle {\begin{aligned}b_{3k}&=2+{\tfrac {1}{3^{3k}}}=2+{\tfrac {1}{27^{k}}}{\overset {k\to \infty }{\to }}2+0=2\\b_{3k+1}&=3+{\tfrac {3k+3}{3k+1}}=3+1+{\tfrac {2}{3k+1}}{\overset {k\to \infty }{\to }}4+0=4,\\b_{3k+2}&=3{\overset {k\to \infty }{\to }}3.\end{aligned}}}

Hence, ${\displaystyle (b_{n})}$ only has the accumulation points ${\displaystyle 2,3}$ and ${\displaystyle 4}$. Furthermore, ${\displaystyle (b_{n})}$ is bounded, as all subsequences are bounded as well. We get ${\displaystyle \liminf _{n\to \infty }b_{n}=2}$ and ${\displaystyle \limsup _{n\to \infty }b_{n}=4}$.

Exercise (Lim inf and Lim sup 2)

1. Let ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ be a sequence of positive real numbers. Prove the following inequalities:
${\displaystyle \liminf _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}\leq \liminf _{n\to \infty }{\sqrt[{n}]{a_{n}}}\leq \limsup _{n\to \infty }{\sqrt[{n}]{a_{n}}}\leq \limsup _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}}$
2. Let ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ be a sequence with
${\displaystyle a_{n}={\begin{cases}{\frac {1}{2^{n}}}&{\text{ für }}n=2k-1,\\{\frac {1}{3^{n}}}&{\text{ für }}n=2k\end{cases}}}$

Compute ${\displaystyle \liminf _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}}$, ${\displaystyle \liminf _{n\to \infty }{\sqrt[{n}]{a_{n}}}}$, ${\displaystyle \limsup _{n\to \infty }{\sqrt[{n}]{a_{n}}}}$ and ${\displaystyle \limsup _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}}$.

3. Derive the following statement from 1): Let ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ be a sequence of positive real numbers with ${\displaystyle \lim \limits _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}=a\in \mathbb {R} }$. Then we have ${\displaystyle \lim \limits _{n\to \infty }{\sqrt[{n}]{a_{n}}}=a}$.
4. Prove the following limits:
${\displaystyle \lim _{n\to \infty }{\frac {1}{\sqrt[{n}]{n!}}}=0\ {\text{ und }}\lim _{n\to \infty }{\frac {n}{\sqrt[{n}]{n!}}}=e}$

Solution (Lim inf and Lim sup 2)

1. Let ${\displaystyle (a_{n})\,}$ be a positive sequence with
${\displaystyle \alpha =\liminf {\frac {a_{n+1}}{a_{n}}}\quad ,\quad \alpha '=\liminf {\sqrt[{n}]{a_{n}}}\quad ,\quad \beta '=\limsup {\sqrt[{n}]{a_{n}}}\quad ,\quad \beta =\limsup {\frac {a_{n+1}}{a_{n}}}}$

We need to show the following inequalities:

${\displaystyle 0\leq \alpha \leq \alpha '\leq \beta '\leq \beta \leq \infty }$

W.l.o.g. let ${\displaystyle \alpha >0\,}$ and ${\displaystyle \beta <\infty }$. By definition, ${\displaystyle \alpha }$ is the smallest and ${\displaystyle \beta }$ the largest accumulation point of the sequence ${\displaystyle \left({\frac {a_{n+1}}{a_{n}}}\right)_{n\in \mathbb {N} }}$. Hence, for all ${\displaystyle \varepsilon >0}$ (${\displaystyle <\alpha \,}$) there exists an ${\displaystyle m\in \mathbb {N} }$ with

${\displaystyle \alpha -\varepsilon <{\frac {a_{k+1}}{a_{k}}}<\beta +\varepsilon \qquad \forall k\geq m}$

Applying the inequality ${\displaystyle (n-m)}$ times setting ${\displaystyle k=m,\ldots ,n-1}$, for an arbitrary ${\displaystyle n\in \mathbb {N} }$ with ${\displaystyle n-1\geq m}$ holds that:

${\displaystyle (\alpha -\varepsilon )^{n-m}<{\frac {a_{m+1}}{a_{m}}}\cdot {\frac {a_{m+2}}{a_{m+1}}}\cdot \ldots \cdot {\frac {a_{n-1}}{a_{n-2}}}\cdot {\frac {a_{n}}{a_{n-1}}}<(\beta +\varepsilon )^{n-m}}$

The expression in the middle is a telescoping product. Except for the numerator ${\displaystyle a_{n}}$ and the denominator ${\displaystyle a_{m}}$, everything cancels out.

${\displaystyle (\alpha -\varepsilon )^{n-m}<{\frac {a_{n}}{a_{m}}}<(\beta +\varepsilon )^{n-m}}$

Multiplying ${\displaystyle a_{m}}$ and taking the ${\displaystyle n}$-th root leads to

${\displaystyle {\sqrt[{n}]{a_{m}}}\cdot (\alpha -\varepsilon )^{1-{\frac {m}{n}}}<{\sqrt[{n}]{a_{n}}}<{\sqrt[{n}]{a_{m}}}\,(\beta +\varepsilon )^{1-{\frac {m}{n}}}}$

With ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{a_{m}}}=1}$, the term on the left side converges to ${\displaystyle \alpha -\varepsilon }$ and the one on the right side converges to ${\displaystyle \beta +\varepsilon }$ for ${\displaystyle n\to \infty }$. Therefore, we get

${\displaystyle \alpha -\varepsilon \leq \liminf {\sqrt[{n}]{a_{n}}}}$ und ${\displaystyle \limsup {\sqrt[{n}]{a_{n}}}\leq \beta +\varepsilon }$

As ${\displaystyle \varepsilon \,}$ can be chosen arbitrarily small, we get

${\displaystyle \alpha \leq \alpha '}$     and     ${\displaystyle \beta '\leq \beta }$

As the inequality ${\displaystyle \alpha '\leq \beta '}$ holds by definition, the claim follows.

• For all ${\displaystyle n\in \mathbb {N} }$ we have ${\displaystyle {\frac {a_{n+1}}{a_{n}}}\geq 0}$, as the numerator and denominator are always positive. Hence, 0 is a lower bound of the quotient sequence ${\displaystyle \left({\frac {a_{n+1}}{a_{n}}}\right)}$. With ${\displaystyle n=2k-1}$ being odd and ${\displaystyle n+1=2k}$ even, it follows that
${\displaystyle {\frac {a_{n+1}}{a_{n}}}={\frac {a_{2k}}{a_{2k-1}}}={\frac {\frac {1}{3^{2k}}}{\frac {1}{2^{2k-1}}}}={\frac {2^{2k-1}}{3^{2k}}}={\frac {2^{2k}}{2\cdot 3^{2k}}}={\frac {1}{2}}\cdot {\frac {4^{k}}{9^{k}}}={\frac {1}{2}}\cdot \left({\frac {4}{9}}\right)^{k}\to 0}$ for ${\displaystyle k\to \infty }$

Hence, there exists a subsequence of the quotient sequence converging to zero. It follows that ${\displaystyle \liminf _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}=0}$

• With ${\displaystyle n=2k}$ being even and ${\displaystyle n+1=2k+1}$ being odd, we get
${\displaystyle {\frac {a_{n+1}}{a_{n}}}={\frac {a_{2k+1}}{a_{2k}}}={\frac {\frac {1}{2^{2k+1}}}{\frac {1}{3^{2k}}}}={\frac {3^{2k}}{2^{2k+1}}}={\frac {3^{2k}}{2\cdot 2^{2k}}}={\frac {1}{2}}\cdot {\frac {9^{k}}{4^{k}}}={\frac {1}{2}}\cdot \left({\frac {9}{4}}\right)^{k}\to \infty }$ for ${\displaystyle k\to \infty }$

Hence, the quotient series doesn't have an upper bound and ${\displaystyle \limsup _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}=\infty }$

• With ${\displaystyle n=2k-1}$ being odd, we have
${\displaystyle {\sqrt[{n}]{a_{n}}}={\sqrt[{2k-1}]{a_{2k-1}}}={\sqrt[{2k-1}]{\frac {1}{2^{2k-1}}}}={\frac {1}{2}}}$

and ${\displaystyle n=2k}$ being even leads to

${\displaystyle {\sqrt[{n}]{a_{n}}}={\sqrt[{2k}]{a_{2k}}}={\sqrt[{2k}]{\frac {1}{3^{2k}}}}={\frac {1}{3}}}$

Hence, ${\displaystyle ({\sqrt[{n}]{a_{n}}})}$ has the accumulation points ${\displaystyle {\frac {1}{2}}}$ and ${\displaystyle {\frac {1}{3}}}$. It follows that

${\displaystyle \liminf _{n\to \infty }{\sqrt[{n}]{a_{n}}}={\frac {1}{3}}}$ and ${\displaystyle \limsup _{n\to \infty }{\sqrt[{n}]{a_{n}}}={\frac {1}{2}}}$
2. As ${\displaystyle \lim _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}=a}$, we get
${\displaystyle \liminf _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}=\limsup _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}=a}$

Using 1) and Der Sandwichsatz, we have

${\displaystyle \liminf _{n\to \infty }{\sqrt[{n}]{a_{n}}}=\limsup _{n\to \infty }{\sqrt[{n}]{a_{n}}}=a}$

It follows that ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{a_{n}}}=a}$.

3. 1st limit: Let ${\displaystyle a_{n}={\frac {1}{n!}}}$. It follows that
{\displaystyle {\begin{aligned}\lim _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}&=\lim _{n\to \infty }{\frac {\frac {1}{(n+1)!}}{\frac {1}{n!}}}\\[0.3em]&=\lim _{n\to \infty }{\frac {n!}{(n+1)!}}\\[0.3em]&=\lim _{n\to \infty }{\frac {n!}{(n+1)n!}}\\[0.3em]&=\lim _{n\to \infty }{\frac {1}{n+1}}=0\end{aligned}}}

Using part 3, we have ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{a_{n}}}=\lim _{n\to \infty }{\sqrt[{n}]{\frac {1}{n!}}}=\lim _{n\to \infty }{\frac {1}{\sqrt[{n}]{n!}}}=0}$.

2nd limit: Let ${\displaystyle a_{n}={\frac {n^{n}}{n!}}}$. It follows that

{\displaystyle {\begin{aligned}\lim _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}&=\lim _{n\to \infty }{\frac {\frac {(n+1)^{n+1}}{(n+1)!}}{\frac {n^{n}}{n!}}}\\[0.3em]&=\lim _{n\to \infty }{\frac {(n+1)^{n+1}n!}{n^{n}(n+1)!}}\\[0.3em]&=\lim _{n\to \infty }{\frac {(n+1)^{n}(n+1)n!}{n^{n}(n+1)n!}}\\[0.3em]&=\lim _{n\to \infty }{\frac {(n+1)^{n}}{n^{n}}}\\[0.3em]&=\lim _{n\to \infty }\left({\frac {n+1}{n}}\right)^{n}\\[0.3em]&=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}=e\end{aligned}}}

Again using 3., we get ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{a_{n}}}=\lim _{n\to \infty }{\sqrt[{n}]{\frac {n^{n}}{n!}}}=\lim _{n\to \infty }{\frac {n}{\sqrt[{n}]{n!}}}=e}$.

Hint

Take a look at the inequalities proven in subtask 1. In subtask 2 even holds that

${\displaystyle \liminf _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}<\liminf _{n\to \infty }{\sqrt[{n}]{a_{n}}}<\limsup _{n\to \infty }{\sqrt[{n}]{a_{n}}}<\limsup _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}}$

Hint

Taking a look at the limits from subtask 4 and using reciprocals, we get:

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{n!}}=\infty \ {\text{ und }}\lim _{n\to \infty }{\frac {\sqrt[{n}]{n!}}{n}}={\frac {1}{e}}}$

## Exercises: Cauchy sequences

Exercise (Cauchy criterion for sequences)

1. Let ${\displaystyle 0 and ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ be a sequence of real numbers with ${\displaystyle |a_{n+1}-a_{n}|\leq q\cdot |a_{n}-a_{n-1}|}$ for all ${\displaystyle n\in \mathbb {N} }$. Show that the sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ converges.
2. Let ${\displaystyle a_{0}=0,a_{1}=1}$ and ${\displaystyle a_{n+1}={\tfrac {1}{2}}(a_{n}+a_{n-1})}$. Show that the sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ converges and compute its limit.

Solution (Cauchy criterion for sequences)

Subtask 1: We will show that ${\displaystyle (a_{n})}$ is a Cauchy sequence. The Cauchy criterion then states that ${\displaystyle (a_{n})}$ converges. Using the assumptions made, we have:

${\displaystyle |a_{j+1}-a_{j}|\leq q\cdot |a_{j}-a_{j-1}|\leq q^{2}\cdot |a_{j-1}-a_{j-2}|\leq \ldots \leq q^{j}\cdot |a_{1}-a_{0}|.}$

If ${\displaystyle m,n\in \mathbb {N} }$ and ${\displaystyle m>n}$, there exists a ${\displaystyle k\in \mathbb {N} }$ with ${\displaystyle m=n+k}$. It follows that

{\displaystyle {\begin{aligned}|a_{m}-a_{n}|&=|a_{n+k}-a_{n}|\\[0.5em]&=|a_{n+k}-a_{n+k-1}+a_{n+k-1}-a_{n+k-2}+\ldots +a_{n+2}-a_{n+1}+a_{n+1}-a_{n}|\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{generalized triangle inequality}}\right.}\\[0.5em]&\leq |a_{n+k}-a_{n+k-1}|+|a_{n+k-1}-a_{n+k-2}|+\ldots +|a_{n+2}-a_{n+1}|+|a_{n+1}-a_{n}|\\[0.5em]&=\sum _{j=0}^{k-1}|a_{n+j+1}-a_{n+j}|\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{index shift}}\right.}\\[0.5em]&=\sum _{j=n}^{n+k-1}|a_{j}-a_{j-1}|\end{aligned}}}

From the inequalities follows that

{\displaystyle {\begin{aligned}|a_{m}-a_{n}|&\leq \sum _{j=n}^{n+k-1}|a_{j}-a_{j-1}|\\[0.5em]&\leq \sum _{j=n}^{n+k-1}q^{j}\cdot |a_{1}-a_{0}|\\[0.5em]&=|a_{1}-a_{0}|q^{n}\sum _{j=0}^{k-1}q^{j}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{geometric sum formula}}\right.}\\[0.5em]&=|a_{1}-a_{0}|q^{n}{\frac {1-q^{k}}{1-q}}\\[0.5em]&\left\downarrow \ q^{k}\geq 0\Rightarrow 1-q^{k}\leq 1\right.\\[0.5em]&\leq {\tfrac {|a_{1}-a_{0}|}{1-q}}q^{n}\end{aligned}}}

From the archimedean axiom we can derive that for all ${\displaystyle \epsilon >0}$ there exists an ${\displaystyle N\in \mathbb {N} }$ such that for all ${\displaystyle n,m\geq N}$ holds

${\displaystyle |a_{m}-a_{n}|\leq {\tfrac {|a_{1}-a_{0}|}{1-q}}q^{n}<\epsilon }$

Hence, ${\displaystyle (a_{n})}$ is a Cauchy sequence and therefore convergent.

Subtask 2: For the sequence ${\displaystyle (a_{n})}$ holds true that

${\displaystyle |a_{n+1}-a_{n}|=|{\tfrac {1}{2}}(a_{n}+a_{n-1})-a_{n}|=|{\tfrac {1}{2}}a_{n-1}-{\tfrac {1}{2}}a_{n}|={\tfrac {1}{2}}|a_{n}-a_{n-1}|.}$

With ${\displaystyle q={\tfrac {1}{2}}}$, the criterion from subtask 1 is met. Hence, ${\displaystyle (a_{n})}$ converges.

The limit can be computed as follows: Analog to subtask 1 - but without taking the absolute values -, we have

${\displaystyle a_{n+1}-a_{n}={\tfrac {1}{2}}(a_{n}+a_{n-1})-a_{n}={\tfrac {1}{2}}(a_{n-1}-a_{n})=-{\tfrac {1}{2}}(a_{n}-a_{n-1}).}$

Applying the formula ${\displaystyle n}$ times leads to

${\displaystyle a_{n+1}-a_{n}=\left(-{\tfrac {1}{2}}\right)^{n}(a_{1}-a_{0}).}$

Hence, we get

${\displaystyle a_{n+1}-\underbrace {a_{0}} _{=0}{\underset {\text{sum}}{\overset {\text{telescoping}}{=}}}\sum _{k=0}^{n}(a_{k+1}-a_{k}){\underset {\text{above}}{\overset {\text{see}}{=}}}\sum _{k=0}^{n}\left(-{\tfrac {1}{2}}\right)^{k}\underbrace {(a_{1}-a_{0})} _{=1-0=1}{\underset {\text{sum formula}}{\overset {\text{geometric}}{=}}}{\tfrac {1-\left(-{\tfrac {1}{2}}\right)^{n+1}}{1+{\tfrac {1}{2}}}}={\tfrac {1-\left(-{\tfrac {1}{2}}\right)^{n+1}}{\tfrac {3}{2}}}={\tfrac {2}{3}}\cdot (1-\left(-{\tfrac {1}{2}}\right)^{n+1}).}$

Using the computation rules for limits, we have

${\displaystyle \lim _{n\to \infty }a_{n+1}=\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }{\tfrac {2}{3}}\cdot (1-\left(-{\tfrac {1}{2}}\right)^{n+1})={\tfrac {2}{3}}\cdot (1-\lim _{n\to \infty }\left(-{\tfrac {1}{2}}\right)^{n+1})={\tfrac {2}{3}}\cdot (1-0)={\tfrac {2}{3}}}$