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I try to solve this question but I don't know how.

given $ a_0 = \frac12 $ and for each $n\geq 1$: $$ |a_n-a_{n-1}| < \frac{1}{2^{n+1}} $$

show that $\{a_n\}$ converges and the limit is $a$ such that $0<a<1$

Update (Edited):

I showed by cauchy that $ |a_m-a_n| < |a_m-a_{m-1}+a_{m+1}-...+a_{n+1}-a_n| < \frac{1}{2^{m+1}} + \frac{1}{2^{m}}+...+\frac{1}{2^{n+2}}$

by the sum of Geometric series, $q=2, a_1=\frac{1}{2^{m+1}}$ then $s_n=\frac{1}{2^{m+1}}[\frac{2^{m-n}-1}{2-1}]$, so $$\frac{1}{2^{m+1}} + \frac{1}{2^{m}}+...+\frac{1}{2^{n+2}} = \frac{1}{2^{m+1}}[\frac{2^{m-n}-1}{2-1}] = \frac{1}{2^{n+1}}-\frac{1}{2^{m+1}}\leq\frac{1}{2^{n+1}}$$

now, it converges!

Can someone help me please to show that the limit is $a$ with $0<a<1$?

Thank you!

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You can prove that $(a_n)$ is Cauchy. – Sangchul Lee Nov 24 '12 at 15:13
but what $a_n$ is? – Alon Shmiel Nov 24 '12 at 15:15
By series what do you mean? – Sangchul Lee Nov 24 '12 at 15:17
Try to estimate $|a_n-a_m| = |a_n-a_{n-1}+a_{n-1}-...-a_m|$. – copper.hat Nov 24 '12 at 15:17
Thanks, Alon. Let me know if the edit/update is correct. – amWhy Nov 25 '12 at 2:46
up vote 2 down vote accepted

Hint: Fix any $m$, and then use triangle inequality and induction to show that $$|a_n-a_m|<\frac1{2^{m+1}}-\frac1{2^{n+1}}$$ for all $n>m$. It follows from this that $\{a_n\}$ is Cauchy, and so converges, say to $a$. In particular, setting $m=0$, we have $$|a_n-a_0|<\frac12-\frac1{2^{n+1}}$$ for all $n$, and since $a_0=\frac12$, we have $0<a_n<1$ for all $n>0$. Thus, $0\leq a\leq 1$.

It remains only to show that $\{a_n\}$ cannot converge to $0$ or to $1$. Note that if we set $b_n=1-a_n$, then $\{b_n\}$ has all the same characteristics as $\{a_n\}$. If we can show that $a\neq 0$, then identical arguments show $b\neq 0$, and so $a=1-b\neq 1$, completing the proof. Thus, we need only show that $a>0$. I recommend noting that $a_1=\frac14+c$ for some $c>0$, and use the work above to conclude that $a\geq c$.

Addendum: Don't waste your time trying to determine the precise values of all the $a_n$s, nor of $a$--there simply isn't enough information given for us to determine this. Fortunately, we don't need that much information to prove the desired results.

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Hint: $\vert a_n - a_0 \vert = \vert a_n - a_{n-1} + a_{n-1} - \dots - a_1 + a_1 - a_0 \vert \leq \sum_{k=1}^n \vert a_k - a_{k-1} \vert$

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You can also prove this generalization the same way: if $|a_{n+1}-a_n| < c_n$ where $c_n$ is decreasing and $\sum_{n=1}^{\infty} c_n$ converges, then $\lim_{n \to \infty} a_n$ exists and is less than $a_1+\sum_{n=1}^{\infty} c_n$.

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thank you Marty, but what's about: 0 < a < 1 ? does it show me that? – Alon Shmiel Nov 25 '12 at 3:39

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