How can I formally proof that the series $\sum^\infty_{n=2} \frac{2}{n^{2}\ln{n}}$ converges?
I am thinking along the lines that because $\sum_{n=2}^{\infty} \frac{2}{n^{2}}$ by the p-series test, then $\sum^\infty_{n=2} \frac{2}{n^{2}\ln{n}}$ must also converge because the fractions I'll be adding would be even smaller. I need to be more formal about it and prove it via some test.
For all $n \ge 2$, we have
$$\ln{n} \ge \frac{1}{2}$$
Hence
$$\frac{2}{n^2 \ln{n}} = \frac{2}{n^2} \frac{1}{\ln{n}} \le \frac{2}{n^2} \frac{2}{1} = \frac{4}{n^2}$$
But then we have
$$\sum_{n = 2}^{\infty} \frac{4}{n^2}$$ converges, since it's a multiple of a $p$-series with $p > 1$.
Use this good fact that if $$\lim_{n\to\infty}~n^pu_n=A$$ and $p>1$ and $A$ is finite. Here $p=2$.
The only test you need is that if $0 < a_n < b_n$ and $\sum b_n$ converges then $\sum a_n$ converges.
Set $a_n = \frac{2}{n^2 \ln n}$ and $b_n = \frac{2}{n^2}$.
Probably overkill, but notice $a(2^n) = \frac{1}{\ln 2 n 2^{2n -1}} \leq \frac{1}{2^{2n -1}}$. So RHS is a geometric series, by comparison test, it converges.
