Show that the sequence $a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$ is upper bounded Let $\{a_{n}\}$ be defined with $a_{1}\in(0,1)$, and 
$$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$$
for all $n\gt 0$.  Show that the sequence is  upper bounded.
My idea: since 
$$a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)$$
then
$$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}}-\dfrac{1}{a_{n}+n^2}$$
then
$$\dfrac{1}{a_{n}}-\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}+n^2}$$
so
$$\dfrac{1}{a_{1}}-\dfrac{1}{a_{n+1}}=\sum_{i=1}^{n}\dfrac{1}{a_{i}+i^2}$$
since
$$a_{n+1}>a_{n}\Longrightarrow \dfrac{1}{a_{i}+i^2}<\dfrac{1}{a_{1}+i^2}$$
so
$$\dfrac{1}{a_{n+1}}>\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)$$
But the RHS might be $\lt0$ for a sufficiently large starting value; for instance, with $a_{1}=\dfrac{99}{100}$
then
$$\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)<0,n\to\infty$$
see:

so this method won't let me bound the series and I don't know what else to do.

We have
$$\frac{a_{n+1}}{a_n} = 1 + \frac{a_n}{n^2}$$
Therefore, if the product
$$\prod_{n=1}^{\infty} ( 1+ \frac{a_n}{n^2})$$
is bounded, then so is $a_n$. Since $1+x \le e^x$, it is enough to show that $\sum_{n\ge 0} \frac{a_n}{n^2} < \infty$. Now, this would follow from $\frac{a_n}{n} < n^{1-\epsilon}$ for some $\epsilon>0$.
Now, for the sequence $b_n\colon = \frac{a_n}{n}$ we have
$$b_{n+1} = \frac{n b_n + b_n^2}{n+1}$$
so by induction $1 > b=b_1 > b_2 > \ldots > 0$.  Now we have
$$\frac{b_{n+1}}{b_n} = \frac{1+ \frac{b_n}{n} }{1 + \frac{1}{n}}$$
Now, the partial products of the infinite  product
$$\prod_{n=1}^{\infty} \frac{1 + \frac{b}{n}}{1 + \frac{1}{n}}$$
decrease like $n^{b-1}$. We are done.
$\bf{Added:}$ A similar argument shows that the sequence of holomorphic function $\phi_1(z) =z$, $\phi_{n+1}(z) = \phi_n(z) + (\frac{\phi_n(z)}{n})^2$,  converges uniformly on compacts to a holomorphic function $\phi(z)$ on the unit disk $D(0, 1)$.
