Does this series converge or diverge? $\sum_{n=1}^\infty\ln\left(1+\frac1{n^2}\right)$ I have a series here, and I'm supposed to determine whether it converges or diverges. I've tried the different tests, but I can't quite get the answer.

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

 A: Hint: Recall that $\ln(1+x)\sim x$ for $x\to 0$, and use the fact that $\sum_{n=1}^\infty\frac1{n^2}$ is convergent.
A: By the Mean Value Theorem
$$\ln(1+x)  =  x \cdot\frac{ 1}{1+cx}\leq x  $$
where   $0 < cx  < x$.
Hence  $0\leq\ln(1 + 1/n^2)\leq 1/n^2$  for each $n\geq 1$. Then  $$\sum_{n=1}^{+\infty}\ln\left(1+\frac{1}{n^2}\right)\leq \sum_{n=1}^{+\infty} \frac{1}{n^2}=\frac{\pi^2}{6}.$$
A: An idea: take the function
$$f(x):=\log\left(1+\frac{1}{x^2}\right):$$
$$\lim_{x\to\infty}\frac{f(x)}{\frac{1}{x^2}}\stackrel{\text{l'Hospital}}=\lim_{x\to\infty}\frac{x^2}{x^2+1}=1$$
Thus, the same as above applies for the discrete variable $\,n\,$ instead of $\,x\,$, and there you have the limit comparison test giving you convergence.
A: From the Weierstrass product of the hyperbolic sine function $$\frac{\sinh\left(\pi x\right)}{\pi x}=\prod_{n\geq1}\left(1+\frac{x^{2}}{n^{2}}\right)
 $$ we have $$\frac{\sinh\left(\pi\right)}{\pi}=\prod_{n\geq1}\left(1+\frac{1}{n^{2}}\right)
 $$ hence $$\sum_{n\geq1}\log\left(1+\frac{1}{n^{2}}\right)=\log\left(\prod_{n\geq1}\left(1+\frac{1}{n^{2}}\right)\right)=\color{red}{\log\left(\frac{\sinh\left(\pi\right)}{\pi}\right)}.$$
