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Show that any sequence of positive numbers $(a_n)$ satisfying $$0< \frac{a_{n+1}}{a_n} \leq 1+ \frac{1}{n^2}$$ must converge.

I have tried taking the limit of the inequality which yields that $0 \leq \lim \frac{a_{n+1}}{a_n} \leq 1$. If $\lim \frac{a_{n+1}}{a_n} \lt 1$, then by the ratio test we have $\sum a_n$ converges, thus $a_n \to 0$. I am trying in particular trying to show that if $\frac{a_{n+1}} {a_n} \to 1$, then $a_n$ is Cauchy thus convergent, of which I am having some trouble.

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This looks like homework? If so, please read this post to learn how to ask a homework question. In particular, we strongly urge you to show your work towards a solution indicating where you are stuck. –  Srivatsan Nov 24 '11 at 3:47
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Why are you interested in that question? Is it homework? If so, please share some thoughts of your own about the problem -- what it reminds you of most, what you've tried that didn't work, what you've tried that worked partially ... –  Henning Makholm Nov 24 '11 at 3:48
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@user11135 : You've asked three questions, all of which have received responses, but you haven't accepted any correct answers. –  Dimitrije Kostic Nov 24 '11 at 4:18
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The argument you are trying to apply can't work, as, if it did, it would also work for the sequence $1,2,3,\dots$, which satisfies $0\lt a_{n+1}/a_n\le1+(1/n)$. You must find a way to make use of that $n^2$ term, something that wouldn't work if it were just $n$. –  Gerry Myerson Nov 24 '11 at 4:45
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3 Answers 3

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Let $L = \liminf_{m \rightarrow \infty} a_m$. This will actually be the limit. We first show $L$ is finite. To see this, multiply ${a_{n+1} \over a_n} < 1 + {1 \over n^2} < e^{1 \over n^2}$ together for $1 \leq n < m$. You then get $$a_m \leq e^{\sum_{n=1}^{m-1}{1 \over n^2}}a_1$$ $$< e^{\pi^2 \over 6}a_1$$ Hence the sequence is bounded and $L$ is finite.

Let $\epsilon > 0$. Since $\liminf_{m \rightarrow \infty} a_m = \sup_n \inf_{m \geq n} a_m$, for each $n, \inf_{m \geq n} a_n \leq L$ and we can choose $m$ arbitrarily large such that $a_m \leq L + \epsilon$. Suppose $M > m$.

Then multiplying ${a_{n+1} \over a_n} < 1 + {1 \over n^2} < e^{1 \over n^2}$ together for $M > n \geq m$, you get $$a_M < e^{\sum_{n=m}^{M-1}{1 \over n^2}}a_m$$ $$< e^{\sum_{n=m}^{\infty}{1 \over n^2}}a_m$$ Given $\epsilon > 0$, if $m$ is large enough the sum $\sum_{n=m}^{\infty}{1 \over n^2}$ can be made less than $\epsilon$. Thus for any $M > m$ you have $$a_M < (L + \epsilon) e^{\epsilon}$$ Since $\epsilon$ was arbitrary you therefore have $\limsup_{m \rightarrow \infty}a_m \leq L$. Since $L$ was defined as $\liminf_{m \rightarrow \infty} a_m$, the overall limit exists and equals $L$.

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What if L = $+\infty$ –  user11135 Nov 24 '11 at 9:02
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@user11135: The "what if" can't happen. The inequality used by Zarrax gives $a_n\le a_1 e^{b}$ where $b$ can be taken to be $\pi^2/6$, though that has no importance. So in particular the sequence $(a_n)$ is bounded. –  André Nicolas Nov 24 '11 at 12:53
    
Thanks for the help, one last quick question though I am having trouble understanding the statement "Note we can choose m arbitrarily large such that $a_m \leq L $. As from my understanding, shouldn't it be the other way around? In particular for m arbitrarily large it is instead that we have $a_m \geq L$? Since $L$ is liminf of $a_n$. Thanks for the help. –  user11135 Nov 25 '11 at 1:36
    
I added an explanation for that and also why $L$ is finite. –  Zarrax Nov 25 '11 at 2:18
    
@Zarrax $1+\frac{1}{n^2}=\frac{n^2+1}{n^2} < \frac{n^2}{n^2-1} \,.$. If you use this inequality, the right sides become telescopic products... –  N. S. Nov 25 '11 at 3:54
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Hint: For $m<n$, find a reasonably good estimate for $\dfrac{a_n}{a_m}$. Use this estimate to show that our sequence is Cauchy.

Details: For fun we give an elementary approach to the estimates. Note that if $k \ge 2$, then $$1+\frac{1}{k^2}<\frac{1}{1-\frac{1}{k^2}}=\frac{k^2}{(k-1)(k+1)}.$$ Let $2\le m<n$. Then $$\prod_{k=m}^n \left(1+\frac{1}{k^2}\right)< \prod_{k=m}^n \frac{k^2}{(k-1)(k+1)}.$$ Write down the product of the first few terms on the right-hand side, and observe the beautiful cancellations. A little examination shows that $$\prod_{k=m}^n \frac{k^2}{(k-1)(k+1)}=\frac{mn}{(m-1)(n+1)}<1+\frac{1}{m-1}.\qquad (\ast)$$

Now we complete the proof of convergence. From the inequality in the statement of the problem, we conclude using $(\ast)$ that $$0<\frac{a_n}{a_m}<1+\frac{1}{m-1}, \qquad\text{and therefore}\qquad |a_n-a_m| <\frac{a_m}{m-1}. $$ But from $(\ast)$ we can see that $a_m<2a_2$. So by picking $m$ large enough, and $n>m$, we can make $|a_n-a_m|<\epsilon$ for any preassigned $\epsilon$, that is, the sequence $(a_n)$ is Cauchy.

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Hint: Define $$ b_n=\prod_{k=1}^{n-1}\left(1+\frac{1}{k^2}\right) $$ and note that $\displaystyle\frac{b_{n+1}}{b_n}=1+\frac{1}{n^2}$. Show that $b_n\le e^{\pi^2/6}$ for all $n$. What does this say about $\lim\limits_{n\to\infty}b_n$?

Consider the sequence $c_n=\dfrac{a_n}{b_n}$. What can you say about $\dfrac{c_{n+1}}{c_n}$?

Further Explanation of Hint: $$ \begin{align} \frac{c_{n+1}}{c_n} &=\frac{a_{n+1}}{a_n}\left/\frac{b_{n+1}}{b_n}\right.\\ &=\frac{a_{n+1}}{a_n}\left/\left(1+\frac{1}{n^2}\right)\right.\\ &\le1 \end{align} $$ So $c_n$ is non-increasing and positive.

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