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For the sequence $a_n = 1 + \frac12 + \frac14 + ... + \frac{1}{2^n}$, $n \ge 1$, find a formula $N = N(\epsilon)$ such that for all $\epsilon > 0$ and for all $m,n \ge N(\epsilon)$, $|a_m - a_n| < \epsilon$.

I have tried many things but it's just not working. I know that the condition is the definition of Cauchy sequences. But that didn't help.

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  • $\begingroup$ Have you tried writing what $a_n-a_m$ is? $\endgroup$ – 5xum Feb 10 '14 at 9:04
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    $\begingroup$ You may notice that $|a_m-a_n|<1/2^N$. Thus you can choose $\epsilon$ accordingly. $\endgroup$ – Zuriel Feb 10 '14 at 9:06
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Hint:

\begin{align*} a_n &= 1 + \frac12 + \frac14 + \cdots + \frac{1}{2^n} \\ &= \frac{1}{2^n} (2^n + 2^{n-1} + 2^{n-2} + \cdots + 1) \\ &= \frac{2^{n+1} - 1}{2^n} = 2 - \frac{1}{2^n} \end{align*} so \begin{align*} |a_n - a_m| &= \left| \left( 2 - \frac{1}{2^n} \right) - \left( 2 - \frac{1}{2^m} \right) \right| \\ &= \left| \frac{1}{2^n} - \frac{1}{2^m} \right| \\ &< \frac{1}{2^m} + \frac{1}{2^n} \\ &\le \frac{1}{2^N} + \frac{1}{2^N} = \frac{2}{2^N} \end{align*}

Pick $N(\epsilon)$ so that $\frac{2}{2^N} < \epsilon$.

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  • $\begingroup$ But doesn't N(epsilon) has to be in the form of: N= some equation in terms of epsilon? How would we get there by just knowing that 2/2^n < epsilon and that N>n $\endgroup$ – ZZZZZZzzzzzz Feb 10 '14 at 18:23
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    $\begingroup$ @ZZZZZZzzzzzz Good question; one way is just to pick $N > \frac{2}{\epsilon}$ (such an $N$ exists by the archimedian property) and then we have $$\frac{2}{2^N} < \frac{2}{N} < \frac{2}{2 / \epsilon} = \epsilon.$$ $\endgroup$ – 6005 Feb 10 '14 at 21:36
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    $\begingroup$ @ZZZZZZzzzzzz And just to clarify, $N$ doesn't have to be an "equation" in terms of epsilon necessarily; it just has to be some function of $\epsilon$. So you can just say, let $N$ be some integer such that $N > \frac{2}{\epsilon}$. $\endgroup$ – 6005 Feb 10 '14 at 21:37
  • $\begingroup$ got it! i was thinking something along the lines of: since 2/2^n < epsilon then n > log[base2] (2/epsilon) and since N(epsilon)>= n. Then N(epsilon)>log[base2](2/epsilon) wouldn't that work? $\endgroup$ – ZZZZZZzzzzzz Feb 10 '14 at 23:22
  • $\begingroup$ @ZZZZZZzzzzzz That works too; I was being cautious just in case you hadn't yet defined the function $\log_2$ in the class. In my analysis class we didn't rigorously define $\log_2$ until much after doing Cauchy sequences, and therefore we wouldn't have been allowed to use it. $\endgroup$ – 6005 Feb 10 '14 at 23:27

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