Convergence $\ln\left(1+\frac{1}{n^2}\right)$

Question

I´m asking for some help, since I am stuck with a proof for convergence like follows:

$$\sum\limits_{n=1}^\infty\ln\left(1+\frac1{n^2}\right)$$

I tried to separate it: $$\sum\limits_{n=1}^\infty\ln\left(\frac{n^2+1}{n^2}\right)=\sum\limits_{n=1}^\infty\left(\ln\left(n^2+1\right) - \ln\left(n^2\right)\right)=...$$

But I am still stuck, maybe my idea is quite wrong. I´m looking forward to some help/solutions. Thank you in advance!

Answer 1

$$ \ln\left(1+\frac{1}{n^2}\right) \underset{(+\infty)}{\sim}\frac{1}{n^2} $$ and $\sum_{n \geq 1}^{ }1/n^2$ converges then it is the same for $\sum_{n \geq 1}^{ }\ln\left(1+1/n^2\right)$.

Or using partial sums : $$ \sum_{k=1}^{N}\ln\left(1+\frac{1}{k^2}\right) \leq \sum_{k=1}^{N}\frac{1}{k^2} \leq 1+\sum_{k=2}^{N}\frac{1}{k\left(k-1\right)} = 1+ \sum_{k=2}^{N}\left(\frac{1}{k-1}-\frac{1}{k}\right) $$ Can you take it from there ?