Exercises: Subsequences, Accumulation points and Cauchy sequences – Serlo

Exercises: Subsequences

Exercise (Subsequences of the e-sequence)

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

Solution (Subsequences of the e-sequence)

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

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

Subtask 2: With

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

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

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

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

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

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

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

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

Exercises: Accumulation points

Exercise (Accumulation points)

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

How to get to the proof? (Accumulation points)

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

Solution (Accumulation points)

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

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

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

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

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

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

Exercises: Lim inf and Lim sup

Exercise (Lim inf and Lim sup)

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

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

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

Solution (Lim inf and Lim sup)

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

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

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

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

With $n=2k$, we get

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

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

• Regarding $(b_{n})$: We have

${\begin{aligned}b_{3k}&=2+{\tfrac {1}{3^{3k}}}=2+{\tfrac {1}{27^{k}}}{\overset {k\to \infty }{\to }}2+0=2\\b_{3k+1}&=3+{\tfrac {3k+3}{3k+1}}=3+1+{\tfrac {2}{3k+1}}{\overset {k\to \infty }{\to }}4+0=4,\\b_{3k+2}&=3{\overset {k\to \infty }{\to }}3.\end{aligned}}$

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

Exercise (Lim inf and Lim sup 2)

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

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

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

Solution (Lim inf and Lim sup 2)

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

   We need to show the following inequalities:

   $0\leq \alpha \leq \alpha '\leq \beta '\leq \beta \leq \infty$

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

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

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

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

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

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

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

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

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

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

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

   $\alpha \leq \alpha '$     and     $\beta '\leq \beta$

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

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

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

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

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

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

   and $n=2k$ being even leads to

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

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

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

   Using 1) and Der Sandwichsatz, we have

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

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

4. 1st limit: Let $a_{n}={\frac {1}{n!}}$. It follows that
   ${\begin{aligned}\lim _{n\to \infty }{\frac {a_{n+1}}{a_{n}}}&=\lim _{n\to \infty }{\frac {\frac {1}{(n+1)!}}{\frac {1}{n!}}}\\[0.3em]&=\lim _{n\to \infty }{\frac {n!}{(n+1)!}}\\[0.3em]&=\lim _{n\to \infty }{\frac {n!}{(n+1)n!}}\\[0.3em]&=\lim _{n\to \infty }{\frac {1}{n+1}}=0\end{aligned}}$

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

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

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

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

Hint

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

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

Hint

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

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

Exercises: Cauchy sequences

Exercise (Cauchy criterion for sequences)

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

Solution (Cauchy criterion for sequences)

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

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

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

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

From the inequalities follows that

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

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

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

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

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

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

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

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

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

Applying the formula $n$ times leads to

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

Hence, we get

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

Using the computation rules for limits, we have

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