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Examine whether the following series converges or not: \( \displaystyle \sum_{n=2}^{\infty}\left ( \frac{1}{\left(n\ln n\right)^2}-n \right ) \).

The series diverges. To see that we'll work with partial sums:

\begin{align*}
\sum_{n=2}^{\infty} \left [ \frac{1}{n^2 \log^2 n} - n \right ] &= \lim_{N \rightarrow +\infty} \sum_{n=1}^{N} \left [ \frac{1}{n^2 \log^2 n} - n \right ] \\
&= \sum_{n=2}^{\infty} \frac{1}{n^2 \log^2 n } - \lim_{N \rightarrow +\infty} \sum_{n=2}^{N} n \\
&=\sum_{n=2}^{\infty} \frac{1}{n^2 \log^2 n } - \lim_{N \rightarrow +\infty} \frac{N\left ( N+1\right )}{2} +1 \\
&\rightarrow -\infty
\end{align*}

It is known that the series $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^2 \log^2 n }$ converges.

P.S: A possible interesting question is to examine the convergence of the series

$$\mathcal{S} = \sum_{n=2}^{\infty} \left [ \frac{1}{n \log n} - n \right ]$$

I have not worked it out but I rush to say that my calculator says it diverges!