The question is if $\sum_{n=1}^{\infty}n(1-\cos(\frac{1}{n^2}))$ converges or not, according to wolfram it does. d'lambert and cauchy's tests don't seem to be the way to go, so im thinking maybe we can compare it to another sum that we know converges and is larger than the sum above.
But which test to use? The obvious choice is to go with $n^2(1-\cos(\frac{1}{n^2}))$ but now we need to prove that this sum converges and that's really the same problem.
Can anyone think of a good comparison?
Hint: $1- \cos^2 \left(\frac{1}{n^2}\right) \leq 2\cdot \left(\frac{1}{n^4}\right)$. This is true due to the inequality: $1- \cos x = 2\sin^2 \left(\frac{x}{2}\right) \leq 2\left(\frac{x}{2}\right)^2$
Since $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^3}$ converges and $\displaystyle\lim_{n\to\infty}\frac{n(1-\cos\frac{1}{n^2})}{\frac{1}{n^3}}=\lim_{n\to\infty}n^4\left(1-\cos\frac{1}{n^2}\right)=\lim_{t\to 0}\frac{1-\cos t}{t^2}=\frac{1}{2}$,
the series converges by the Limit Comparison Test.
The limit comparison test with $\sum\frac{1}{n^2}$ (and two applications of L'hopital's Rule) should do the trick.
