Convergence of sequence: $f(n+1) = f(n) + \frac{f(n)^2}{n(n+1)}$ I am trying to work out a problem on some old analysis qual exam, and managed to reduce it to this question, but I can't seem to figure out this final step:
Consider the sequence defined by the recurrence relation $$f(n+1) = f(n) + \frac{f(n)^2}{n(n+1)}$$ with initial condition $f(1) = 1$. Does this sequence converge or diverge?

We can answer this question in the same manner as this answer.
Since $f(1)=1$ and
$$
f(n+1)=f(n)+\frac{f(n)^2}{n(n+1)}\tag{1}
$$
we have that $f(n)$ is increasing and inductively we have that
$$
1\le f(n)\le n\tag{2}
$$
Furthermore, $(1)$ implies that
$$
\begin{align}
&\left(\frac1{f(n+1)}-\frac1{n+1}\right)-\left(\frac1{f(n)}-\frac1n\right)\\
&=\left(\frac1n-\frac1{n+1}\right)-\left(\frac1{f(n)}-\frac{n(n+1)}{n(n+1)f(n)+f(n)^2}\right)\\
&=\frac1{n(n+1)}-\left(\frac{n(n+1)+f(n)}{n(n+1)f(n)+f(n)^2}-\frac{n(n+1)}{n(n+1)f(n)+f(n)^2}\right)\\
&=\frac1{n(n+1)}-\frac1{n(n+1)+f(n)}\\
&=\frac{f(n)}{n(n+1)(n(n+1)+f(n))}\\
&\ge\frac{f(n)}{n^2(n+1)(n+2)}\\
&\ge\frac1{n^2(n+1)(n+2)}\tag{3}
\end{align}
$$
Since
$$
\sum_{n=1}^\infty\frac1{n^2(n+1)(n+2)}=\frac{\pi^2}{12}-\frac58\tag{4}
$$
we have
$$
\begin{align}
\frac1{f(n)}
&\ge\frac1n+\sum_{k=1}^{n-1}\frac1{n^2(n+1)(n+2)}\\
&\ge\frac{\pi^2}{12}-\frac58\tag{5}
\end{align}
$$
Therefore,
$$
f(n)\le\frac{24}{2\pi^2-15}\tag{6}
$$
Since $f(n)$ is increasing and bounded above, it converges.
