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.

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.
$$ f(a_{1}) \geq \frac{a_{1}}{1 - a_{1}}. $$
