I'm trying to prove that this series converge through comparison test.
$$\sum _{n=0}^{\infty } \frac{\log(1+n)}{n^2}$$
The problem is that I can't find a series to compare.
For example:
$$\sum _{n=0}^{\infty } \frac{\log(1+n)}{n^2}>\sum _{n=0}^{\infty } \frac{1}{n^2}$$
Since $\ln y < y$ , letting $y=x^s$, We have $$\ln (x^s) < x^s $$
$$\implies \log x < \frac{x^s}{s} \; \forall s \in \Bbb R^+ $$
For $s=\frac 12$ , we get $$\log n <2\sqrt n \implies \log (n+1) < 2\sqrt{n+1}$$
Therefore $$\sum \frac{\log (1+n)}{n^2} < \sum \frac{ 2\sqrt {n+1}}{n^2} \sim_\infty \frac{2}{n^{3/2}} $$
Now since $\frac{2}{n^{3/2}} =2\zeta \left( \frac 32 \right) $ and $\zeta (s) $ converges for all $s >1$, $\displaystyle \sum \frac{\log (1+n)}{n^2} $ is convergent.
The trick is to compare by limit comparison test with
$$\sum _{n=0}^{\infty } \frac{1}{n^{\frac32}}$$
or more in general with
$$\sum _{n=0}^{\infty } \frac{1}{n^{a}}$$
with $1<a<2$.
By comparison test we have $$\frac{n^{3/2}\log (n)}{n^2}= \frac{\log (n)}{n^{1/2}} =\left(\frac{\log (n)}{n^{1/2}}\right)=2\frac{\log (n^{1/2})}{n^{1/2}}\to 0$$ that is for n large enough we have
$$\frac{\log (n)}{n^2}<\frac{1}{n^{3/2}}$$
the convergence follow by Riemann series. see more general here On convergence of Bertrand series $\sum\limits_{n=2}^{\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}$ where $\alpha, \beta \in \mathbb{R}$
You need to know that the logarithm grows slower than all positive powers: $$\log n=o(n^\alpha), \alpha>0.$$
Indeed, when you increase $n$ geometrically, the LHS grows like an arithmetic sequence, while the RHS like a geometric one.
Then your general term can be bounded by some $n^{\alpha-2}<n^{-1}.$
$\int_{1}^{1+n}x^{-1/2}dx = $
$2(n+1)^{1/2} -2(1^{1/2})< 2(n+1)^{1/2} $.
Also we have:
$2(n+1)^{1/2} > \int_{1}^{1+n}(1/x^{1/2})dx \gt$
$ \int_{1}^{1+n}(1/x) dx = \ln(1+n)$.
Hence:
$2(n+1)^{1/2} \gt \ln(1+n)$.
Ready for the comparison test.
