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.
 A: Now,today I have solve this problem,I post my methods,I hope is not wrong?
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
$$\dfrac{1}{a_{1}}-\dfrac{1}{a_{n+1}}=\sum_{i=1}^{n}\dfrac{1}{a_{i}+i^2}$$
so
$$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{1}}-\left(\dfrac{1}{a_{1}+1^2}
+\dfrac{1}{a_{2}+2^2}+\cdots+\dfrac{1}{a_{n}+n^2}\right)$$
 by induction,we have
 $$a_{n}>nt^{n+1},t=\sqrt{a_{1}}\in (0,1)$$
 because
 $$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}>nt^{n+1}+t^{2n+2}$$
 we only prove
 $$nt^{n+1}+t^{2n+2}>(n+1)t^{n+2},t=\sqrt{a_{1}}\in (0,1)$$
 $$\Longleftrightarrow n+t^{n+1}-(n+1)t>0$$
 use this Bernoulli inequality:
 $$(1+x)^n\ge 1+nx,x>-1,n>1$$
 then we have
 $$t^{n+1}=(1+t-1)^{n+1}>1+(n+1)(t-1)=1+(n+1)t-(n+1)$$
 Now $$\dfrac{1}{a_{n+1}}>\dfrac{1}{t^2}-\left(\dfrac{1}{t^2+1}+\dfrac{1}{2t^3+2^2}+\cdots+\dfrac{1}{nt^{n+1}+n^2}\right)$$
 other hand,Use AM-GM inequality,we have
 \begin{align*}
 \dfrac{1}{t^2+1}+\dfrac{1}{2t^3+2^2}+\cdots+\dfrac{1}{nt^{n+1}+n^2}&=\dfrac{1}{1(1+t^2)}+\dfrac{1}{2(1+1+t^3)}+\cdots+\dfrac{1}{n(1+1+\cdots+1+t^{n+1})}\\
 &<\dfrac{1}{2t}+\dfrac{1}{2\times 3t}+\cdots+\dfrac{1}{n(n+1)t}\\
 &=\dfrac{1}{t}\left(1-\dfrac{1}{n+1}\right)\\
 &<\dfrac{1}{t}
 \end{align*}
 so
 $$\dfrac{1}{a_{n+1}}>\dfrac{1}{t^2}-\dfrac{1}{t}$$
 $$\Longrightarrow a_{n+1}<\dfrac{t^2}{1-t}$$
A: Here is a solution: Since $(a_{n})$ is monotone increasing, either it remains bounded or it diverges to $+\infty$. Assume that $a_{n} \nearrow +\infty$.
Observation 1. Referring to OP's identity
$$ \frac{1}{a_{n}} - \frac{1}{a_{n+1}} = \frac{1}{a_{n} + n^{2}}, $$
we have
$$ \frac{1}{a_{n}} - \frac{1}{a_{m+1}} = \sum_{k=n}^{m} \frac{1}{a_{k} + k^{2}}. $$
Taking $m \to \infty$, it follows that
\begin{align*}
\frac{1}{a_{n}}
= \sum_{k=n}^{\infty} \frac{1}{a_{k} + k^{2}}
\leq \sum_{k=n}^{\infty} \frac{1}{k^{2} - k}
= \frac{1}{n-1}.
\end{align*}
Thus we must have $n-1 \leq a_{n}$ for any $n \geq 2$.
Observation 2. Now we prove that $a_{n} \leq a_{1} n$ for any $n$. Indeed, this is trivial when $n = 1$. Also, if it holds for $n$ then
$$ a_{n+1} = a_{n} \left( 1 + \frac{a_{n}}{n^{2}} \right) \leq n a_{1} \left( 1 + \frac{1}{n} \right) = (n+1)a_{1}. $$
Therefore by induction, we have the desired estimate.
Conclusion. Combining two observation, it follows that
$$n - 1 \leq a_{n} \leq na_{1} \quad \Longrightarrow \quad 1 - \frac{1}{n} \leq a_{1}.$$
for $n \geq 2$. Taking $n\to\infty$, we have $1 \leq a_{1}$, a contradiction! Therefore $(a_{n})$ must remain bounded.

Addendum. Now let us investigate an asymptotic behavior of the limit $f(a_{1}) = \lim_{n\to\infty} a_{n}$. We have
\begin{align*}
\frac{1}{a_{1}} - \frac{1}{f(a_{1})}
&= \sum_{n=1}^{\infty} \frac{1}{n^{2} + a_{n}}
\geq \sum_{n=1}^{\infty} \frac{1}{n^{2} + n}
= 1,
\end{align*}
So we have a lower bound.
$$ f(a_{1}) \geq \frac{a_{1}}{1 - a_{1}}. $$
A: 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 $D(0, 1)$.
