Prove or disprove the recursively defined sequence is convergent. The sequence $\{a_n\}$ is defined by $a_1=1, a_2=0$ and $a_{n+2}=a_{n+1}+\displaystyle\frac{a_n}{n^2}$ for $n\in \mathbb{N}$.
Since $\displaystyle\frac{1}{n^2}$ is summable, when $n$ is large, the sequence is something like $a_n=a_{n-1}+\displaystyle\sum_{i\leq n-2}\frac{a_i}{i^2}$, so I think the sequence should be convergent.
Then I want to use the Monotone convergent theorem, i.e. to show $\{a_n\}$ is monotonic and bounded.
For monotonic, it is easy to see that $\{a_n\}$ is increasing.
But for the upper bound, assuming $\{a_n\}$ converges and taking the limit $n\to \infty$ does not give any hints for me to find a suitable upper bound. I have also used computer programs to compute up to the 10000th term, but it seems that $\{a_n\}$ is still increasing, does not converges to a certain number.
So I wonder if it is convergent or not.
 A: Well this took longer than I thought. I feel like there must be an easier solution...

Claim 1: $a_n\le\sqrt n$ for all $n$. This holds for $n=1$ and $n=2$. Actually, we will want to assume $n\ge 3$ later, so we can also check $a_3=1\le\sqrt3$. Now if $a_n\le\sqrt n$ and $a_{n+1}\le\sqrt{n+1}$, then
$$a_{n+2}=a_{n+1}+\frac{a_n}{n^2}\le\sqrt{n+1}+\frac{\sqrt n}{n^2},$$
and it suffices to show $\sqrt{n+1}+\frac{\sqrt n}{n^2}\le \sqrt{n+2}$. Note
$$\sqrt{n+1}+\frac{\sqrt n}{n^2}\le\sqrt{n+1}+\frac{\sqrt{n+1}}{n^2}=\sqrt{n+1}\left(1+\frac{1}{n^2}\right)$$
and the inequality
$$\sqrt{n+1}\left(1+\frac{1}{n^2}\right)\le\sqrt{n+2}$$
is equivalent to
$$(n+1)\left(1+\frac{1}{n^2}\right)^2\le n+2.$$
With some elbow grease this is equivalent to
$$n^4\ge 2n^3+2n^2+n+1.$$
Now since $n\ge 3$,
$$n^4\ge 3n^3=2n^3+n^3\ge 2n^3+3n^2=2n^3+2n^2+n^2\ge 2n^3+2n^2+n+1.$$
This establishes Claim 1.

Claim 2: $a_n=\sum_{i=1}^{n-2}\frac{a_i}{i^2}$ for $n\ge 3$. This holds for $n=3$, and if $a_n=\sum_{i=1}^{n-2}\frac{a_i}{i^2}$, then
$$a_{n+1}=a_n+\frac{a_{n-1}}{(n-1)^2}=\frac{a_{n-1}}{(n-1)^2}+\sum_{i=1}^{n-2}\frac{a_i}{i^2}=\sum_{i=1}^{n-1}\frac{a_i}{i^2}.$$

Finishing up: we now have $a_n=\sum_{i=1}^{n-2}\frac{a_i}{i^2}\le\sum_{i=1}^{n-2}n^{-\frac32}.$ Pick your favorite way to show this is the partial sum of a convergent $p$-series, and we're done!
A: Hint: Show that $a_n \leq \prod_{i \leq n} (1+\frac  1{i^{3/2}})$ for all $n \geq 4$.  The infinite product $\prod_{i \leq n} (1+\frac  1{i^{3/2}})$ is convergent because $\sum_n \frac 1 {n^{3/2}} <\infty$.
[ Above inequality may hold for $n <4$ also but I found it easy to verify it for $n \geq 4$].
A: Here is yet another approach: Clearly $a_n\ge 0$ for all $n$, so we can derive the following for all $n>1$: $a_{n}\le a_{n+1}$ and therefore $a_{n+2} \le a_{n+1} + a_{n+1}/n^2$, or equivalently
$$
  \frac{a_{n+2}}{a_{n+1}}-1 \le \frac{1}{n^2}.
$$
For all positive $x$, we have $\log x\le x-1$ and so it follows that
$$
  \log\frac{a_{n+2}}{a_{n+1}} \le \frac{1}{n^2}.
$$
Since all the terms $\log(a_{n+2}/a_{n+1})$ are non-negative and are dominated by the sequence $1/n^2$ whose sum converges, the sum of $\log(a_{n+2}/a_{n+1})$ converges too. But its partial sum from $n=2$ to $n=m$ is simply $\log(a_{m+2}/a_{3})$, so $a_{m+2}$ clearly converges.
