Problem : Suppose $a_{n+1}=a_{n} + \frac{a_n^2}{n^2}$, and $ 0 \leq a_1 < 1 $. Prove $lim_{n\to\infty}a_n$ exists.
Here is my current solution, which only holds partially:
If $a_1 = 0$, then $a_n = 0$;
For $a_1 > 0$, $a_{n+1} - a_{n} = \frac{a_n^2}{n^2} \geq 0$, $a_n$ increases monotonically. So
$a_{n+1}=a_{n} + \frac{a_n^2}{n^2} \leq a_{n} + \frac{a_n \cdot a_{n+1}}{n^2}$, divide both sides by ${a_n \cdot a_{n+1}}$, we got $\frac{1}{a_n} - \frac{1}{a_{n+1}} \leq\frac{1}{n^2} $
We can sum both sides: $\sum_{i=1}^{n}(\frac{1}{a_i}-\frac{1}{a_{i+1}}) \leq \sum_{i=1}^{n}\frac{1}{i^2}$, which means $\frac{1}{a_1} - \frac{1}{a_{n+1}}\leq\sum_{i=1}^{n}\frac{1}{i^2}$
When n approaches infinity, the right hand side approaches $\frac{\pi^2}{6}$ (skip this proof)
which means $lim_{n\to\infty}\frac{1}{a_{n+1}} \geq \frac{1}{a_1} - \frac{\pi^2}{6} $
Now here is the tricky part: For $\frac{1}{a_1} > \frac{\pi^2}{6}$, I can take the reciprocal of both sides,
and have $lim_{n\to\infty}{a_{n+1}} \leq \frac{1}{\frac{1}{a_1}-\frac{\pi^2}{6}}$, so the limit exists
However, if $1 < \frac{1}{a_1} \leq \frac{\pi^2}{6}$, I cannot do this. Then does the limit exists in this case? I write a code in MATLAB when $a_1 = \frac{6}{\pi^2}$ and run a very large loop, seems $a_n$ exists and is between 2.2 and 2.3.Is there a way to prove? Thanks.
