Limit of $a_{n+1} = a_{n}(1+1/n^2)$ 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.
 A: 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 $\{a_n\}$.
A: hint
By induction, it is easy to see that
$$(\forall n\in \Bbb N) \;\; a_n>0$$
put $$b_n=\ln(a_n)$$
then
$$b_{n+1}=b_n+\ln(1+\frac{1}{n^2})$$
$$\le b_n +\frac{1}{n^2}$$
A: 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.
A: 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.
A: 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.
