If $\lim_{n \rightarrow \infty} (a_{n+1}-\frac{a_n}{2})=0$ then show $a_n$ converges to $0$. I have been stuck on this question for a while now. I have tried many attempts. Here are two that I thought looked promising but lead to a dead end:  Attempt 1:  Write out the terms of $b_n$:  $$b_1=a_{2}-\frac{a_{1}}{2}$$ $$b_2=a_{3}-\frac{a_{2}}{2}$$ $$b_3=a_{4}-\frac{a_{3}}{2}$$ $$\cdots$$ $$b_n=a_{n+1}-\frac{a_{n}}{2}$$  Adding up the terms you get:  $$\sum_{i = 1}^n b_i=a_{n+1}+\frac{a_n}{2}+\frac{a_{n-1}}{2}+\cdots+\frac{a_2}{2}-\frac{a_1}{2}.$$  But a dead end here.  Attempt 2:  For $ε=\dfrac{1}{2}$, $\exists K$ such that $\forall n&gt;K$, $$\left|a_{n+1}-\frac{a_n}{2}\right|&lt;\frac{1}{2}.$$  Now I attempt to prove $\{a_n\}$ is Cauchy and hence converges.   For $m&gt;n&gt;K$, \begin{align*} |a_m-a_n|&amp;=\left|a_m-\frac{a_{m-1}}{2}+\frac{a_{m-1}}{2}-\frac{a_{m-2}}{2^2}+\cdots -+\frac{a_{n+1}}{2^{m-n-1}}-a_n\right|\\ &amp;\leq \left|a_m-\frac{a_{m-1}}{2}\right|+\frac{1}{2}\left|a_{m-1}-\frac{a_{m-2}}{2}\right|+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\ &amp;\leq \frac{1}{2}+\frac{1}{2} × \frac{1}{2}+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\ &amp;&lt;1+\left|\frac{a_n}{2^{m-n}}-a_n\right|, \end{align*} and a dead end.
Show that $x_{n+1}-\dfrac{x_n}{2}$ converges to zero implies that $(x_n)_n$ also converges to zero. Let $(x_n)_n$ be a sequence of real numbers such that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0.$$ Show that $(x_n)_n$ must also converge to zero.     $\textbf{My Attempt:}$   My original idea was to use the triangle inequality and the fact that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0$$   to show that the sequence $(x_n)_n$ is Cauchy and hence converges to some real number $a$, which would give me $$\lim\limits_{n\rightarrow \infty}x_n=a=\lim\limits_{n\rightarrow \infty}x_{n+1}\implies a-\dfrac a 2= \lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0\implies a=0.$$ However, I'm having trouble showing that $(x_n)_n$ is Cauchy.   Any suggestions/hints would be greatly appreciated. Thanks!
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