I solved this series $$\sum_{n=2}^\infty \frac 1 {n^2\ln n}$$ wit Condensation Test and I got now $$\sum_{n=2}^\infty \frac{1}{2^nnln(ln(2))}$$
Can I use now a geometric series like $\sum_{n=0}^\infty \frac 1 {2^n}$ with comparasion test for say that $\frac{1}{2^nln(ln(2))}<\frac 1 {2^n}$ and get the conclusion that the series is Convergent?
Thanks
Yes, factor out the $\frac{1}{\ln \ln 2}$ and the comparison $\frac{1}{2^n n} < \frac{1}{2^n}$ will work, because $$\frac{1}{2^n n} < \frac{1}{2^n} \implies 2^n n > 2^n \implies n > 1$$ and $\sum \frac{1}{2^n}$ is a convergent geometric series.
Maybe it's easier to observe that $\dfrac{1}{n^{2}\sqrt{n}}<\dfrac{1}{n^2 \log (n)}$ for any $n\ge 2$ because $\sqrt n>\log n$ and $\dfrac{1}{n^{2+\frac12}}$ converges
so the given series converges
hope it helps
