Is the sequence $x_n=\frac{1}{n^2} \sum_{k=1}^n \frac{k^2}{(k+1)\log (k+1)}$ convergent to $0$? 
Is it true that $$ \frac{1}{n^2}\sum_{k=1}^n \frac{k^2}{(k+1)\log(k+1)} \to 0$$
  as $n\to \infty$?

It is clear that
$$
\sum_{k=1}^n \frac{k^2}{(k+1)\log(k+1)}\le \sum_{k=1}^n \frac{k}{\log(k+1)}
$$
So if we didn't have that "$\log$" term the statement would be false. But I can't think of a good approximation to the term $k/\log(k+1)$ to proceed further from here. None of the standard theorems I know of seem to apply too.
I highly suspect the sequence does converge to $0$, from the context where it was derived. Any good ideas to proceed from here?
Thanks

Arguing a little sloppily,
where "$\sim$" means
"within a factor of",
$\begin{array}\\
\frac{1}{n^2}\sum_{k=1}^n \frac{k^2}{(k+1)\log(k+1)}
&\sim \frac{1}{n^2}\sum_{k=2}^n \frac{k^2}{k\log(k)}
\qquad\text{since we can ignore }k=1\\
&\sim \frac{1}{n^2}\sum_{k=2}^n \frac{k}{\log(k)}\\
&\sim \frac{1}{n^2\log(n)}\sum_{k=2}^n k
\qquad\text{since }\log k > \frac12 \log n \text{ for }k > \sqrt{n}\\
&\sim \frac{1}{\log(n)}\\
&\to 0\\
\end{array}
$
Therefore it does converge to zero.
