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I read the following exercise :

Let $(a_n)$ be a sequence such as :

$a_1 = 2$, $a_{n+1} = a_n.(1+1/n^2)$

Prove that $\lim_{n \rightarrow \infty} a_n$ exists.

Hint : $\ln : \: (0, \infty) \rightarrow \mathbb{R} \:$ is a continuous, increasing function such as $\ln(1+x) \leq x$ and $\ln(ab) = \ln(a)+\ln(b)$.

It is quite clear that $a_n$ is increasing but then I don't really know how to proceed. Could you please help me ?

Thanks.

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  • $\begingroup$ write $a_{n+1}$ by substituting the value of a(n).... recursively... then, use rienmann sums $\endgroup$ Nov 22, 2020 at 15:37
  • $\begingroup$ @PNDas Yes I agree but how ? And what would the ln function be useful for ? $\endgroup$
    – Laury
    Nov 22, 2020 at 15:37
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    $\begingroup$ If you want to use $\ln$ then see $\ln(a_{n+1})=\ln(a_n)+\ln(1+\frac 1{n^2})\leq\ln(a_n)+\frac 1{n^2}\leq a_{n-1}+\frac 1{{n-1}^2}+\frac 1{n^2}$, inductively, $\ln(a_{n+1})\leq a_1+ \sum_{m=1}^n\frac 1{m^2}\leq a_1+ \sum_{m=1}^{\infty}\frac 1{m^2}= a_1+ \frac {{\pi}^2}6$ $\endgroup$
    – PNDas
    Nov 22, 2020 at 15:45
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    $\begingroup$ @PNDas How does the limit of the ratio being 1 show above boundedness? Take $a_n=n^2$ for example. Limit of the ratio is still 1 but the sequence is unbounded above. Really not following the logic. $\endgroup$ Nov 22, 2020 at 18:44
  • $\begingroup$ @CogitoErgoCogitoSum,yeah you are right, but I didn't say limit of ratio =1 then sequence is bounded. $\endgroup$
    – PNDas
    Nov 23, 2020 at 5:50

5 Answers 5

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As $\ln x$ is increasing, $b_n = \ln a_n$ is an increasing sequence.

Also $$b_{n+1}= \sum_{k=1}^n \ln(1+1/k^2) \le \sum_{k=1}^n 1/k^2$$

according to the provided hint.

As the series on the RHS is convergent, the sequence $\{b_n\}$ is bounded. Being also increasing, it is convergent. Therefore so is $\{a_n\}$.

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Prove by induction that $\ln a_n \le \sum_{k=1}^{n-1}\frac1{k^2}$ for all $n \ge 2$.

For $n=2$ we have $$\ln a_2 = \ln \frac52 < 1$$ and if we assume that $\ln a_n \le \sum_{k=1}^{n-1}\frac1{k^2}$ for some $n$ then $$\ln a_{n+1} = \ln a_n + \ln\left(1+\frac1{n^2}\right) \le \sum_{k=1}^{n-1}\frac1{k^2} + \frac1{n^2} = \sum_{k=1}^n \frac1{k^2}$$ which completes the proof.

Hence $$\ln a_{n} \le \sum_{k=1}^\infty \frac1{k^2} = \frac{\pi^2}6 \implies a_n \le e^{\frac{\pi^2}6}$$ so your sequence is bounded.

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hint

By induction, it is easy to see that $$(\forall n\in \Bbb N) \;\; a_n>0$$ put $$b_n=\ln(a_n)$$

then

$$b_{n+1}=b_n+\ln(1+\frac{1}{n^2})$$ $$\le b_n +\frac{1}{n^2}$$

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Take $\ln$ about both sides, we have $\ln(a_{n+1})=\ln(a_n)+\ln(1+\frac{1}{n^2})\leq\ln(a_n)+\frac{1}{n^2}$.

Then $\ln(a_{n+1})-\ln(a_n)\leq\frac{1}{n^2}$.

adding them above $n=1,2,\cdots$ $$\underset{n\to\infty}{\lim\sup}\ln(a_n)-\ln(a_1)\leq\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$ So,$\underset{n\to\infty}{\lim\sup}\ln(a_n)\leq \ln(a_1)+\frac{\pi^2}{6}$, and $\underset{n\to\infty}{\lim\sup} a_n\leq 2e^{\frac{\pi^2}{6}}$. Therefore, the sequence is bounded. Furthermore, the sequence has a limit point due to increasing.

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  • $\begingroup$ I think it is more interesting to evaluate the limit point. $\endgroup$
    – fengbiqian
    Nov 22, 2020 at 15:58
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By using $\ln (1+x)\le x$ for $x\ge 0$, we have $$ 1+{1\over n^2}\le e^{1\over n^2} $$ hence $a_n$ is bounded above by another sequence $b_n$ where $$ b_{n+1}=b_ne^{1\over n^2}\quad,\quad b_1=4 $$ Since $b_n\to b_1e^{\sum_{i=1}^\infty {1\over i^2}}=4e^{\pi^2\over 6}$ then $a_n$ also tends to a limit because it is strictly increasing and bounded from above.

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