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Problem:

If $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 0$, show that $\lim_{n\to\infty}a_n = 0$.

Attempted Proof:

Let $\epsilon > 0$. From the hypothesis, $\exists \ N \in \mathbb{P}$ such that if $n \geq N$, then $$\left|\ \frac{a_{n+1}}{a_n} - 0 \ \right| < \epsilon.$$

This implies

$$\left| {a_{n+1}} \right| < \epsilon \left| a_n \right|.$$

Thus, let $n\geq N'$ such that $$\left|\ \frac{a_{n+1}}{a_n} - 0 \ \right| < \epsilon\left|a_n\right|.$$

Then we have $\left|a_{n+1}\right| < \epsilon,$ which implies $\lim_{n\to\infty}a_n = 0$.

My main concern with my proof is that $\epsilon$ depends on $a_n$. Is this an issue?

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    $\begingroup$ You're nearly there with your second equation. If $|a_{n+1}| < \epsilon |a_n|$, for $n \geq N$, can you relate $|a_m|$ to $|a_N|$ for $m\geq N$, maybe by induction? This should give you what you need. $\endgroup$
    – mlk
    Jan 25 '17 at 15:03
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    $\begingroup$ It is very definitely an issue . Write $|a_{n+1}/a_n|<\epsilon_n |a_n|$. This does not imply that $\epsilon_n$ can be arbitrarily small. E.g. if $a_n=2^{-n^2}$ then $\epsilon_n=2^{n^2-2n-1}.$ $\endgroup$ Jan 25 '17 at 16:04
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Since $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=0$, then for $\epsilon=\frac12$, there is $N\in\mathbb{N}$ such that when $n\ge N$, $$ \bigg|\frac{a_{n+1}}{a_n}\bigg|<\epsilon. $$ Thus for $n>N$, one has $$ \bigg|\frac{a_{N+1}}{a_N}\bigg|<\epsilon, \bigg|\frac{a_{N+2}}{a_{N+1}}\bigg|<\epsilon,\cdots,\bigg|\frac{a_{n}}{a_{n-1}}\bigg|<\epsilon$$ and hence $$ \bigg|\frac{a_{n}}{a_N}\bigg|<\epsilon^{n-N}$$ or $$ |a_n|<\epsilon^{n-N}|a_N|. $$ So $$ \lim_{n\to\infty}a_n=0.$$

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  • $\begingroup$ I chose your answer because it uses the information currently available to me in my textbook. $\endgroup$ Jan 25 '17 at 18:09
  • $\begingroup$ From your second to last step, how do we choose the correct $\epsilon$ to show that for $n \geq N$ we have $\left|a_n - 0\right| < \epsilon$? I do see that the limit must be zero, but my professor told me that I should find the largest possible $\epsilon$ to fulfill the inequality (basically using the definition of the limit). $\endgroup$ Jan 25 '17 at 18:24
  • $\begingroup$ Can we choose $0 < \frac{\epsilon}{a_N} < 1$ and then argue that $\left| a_n - 0 \right| < \left(\frac{\epsilon}{\left| a_N \right|}\right)^{n-N} \left| a_N \right| < \frac{\epsilon}{\left|a_N\right|}\left|a_N\right| < \epsilon$? $\endgroup$ Jan 25 '17 at 18:31
  • $\begingroup$ @EternusVia, we can't choose $0<\frac{\epsilon}{a_N}<1$ since we don't have information for $a_N$. $\endgroup$
    – xpaul
    Jan 25 '17 at 20:24
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For $\epsilon =1$, one gets $|a_{n+1}| < |a_n|$ for each $n \ge N$, so as $(|a_n|)$ is bounded below, we deduce that it is convergent. Suppose that it converges to some $L >0$ and get a contradiction by proving that $\lim \frac{a_{n+1}}{a_n}$ would then be $1$. Therefore it converges to $0$ and so does $(a_n)$.

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That would be an issue, yes. The quickest way to prove this is to recognize that $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=0$ tells you that $\sum_0^\infty a_n$ converges by the ratio test. Thus $a_n$ must go to $0$.

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You should study $|a_n|$ and see that after $N$ big enough, you have $|a_{n+1}|<|a_n|$...

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