# Prove $\sum_{k = 2}^\infty \ln(1+\frac{1}{k^2})$ converges using $\exp(x) \geq 1+x$.

All I've got so far is

$$\exp(x) \geq 1+x \Rightarrow x \geq \ln(1+x) \Rightarrow \frac{1}{k^2} \geq \ln\left(1+\frac{1}{k^2}\right)$$

which (since $\ln(1+\frac{1}{k^2})$ is larger than zero) means that

$$\sum_{k = 2}^\infty \ln\left(1+\frac{1}{k^2}\right) < \sum_{k = 2}^\infty \frac{1}{k^2}.$$

But I can't use the comparison test because I don't know how to prove $\sum_{k = 2}^\infty \frac{1}{k^2}$ converges.

• The integral test could also be used, since it is well known that $\int_0^\infty \log(1+1/x^2)\ dx = \pi$. – user1892304 Jun 4 '16 at 17:44

I don't know how to prove $\sum_{k = 2}^\infty \frac{1}{k^2}$ converges.
Hint. You may consider a telescoping sum, $$0<\sum_{k=2}^\infty\frac1{k^2}< \sum_{k=2}^\infty\frac1{k(k-1)}=\sum_{k=2}^\infty\left(\frac1{k-1}-\frac1{k}\right)=1<\infty.$$
One easy way to check that $\sum_{k=2}^\infty\frac1{k^2}$ converges is the integral test: $$\int_2^\infty\frac1{x^2}\,dx=\left.-\frac1x\right|_2^\infty=\frac12.$$
Why not to do more? Since the Weierstrass product for the $\sinh$ function gives: $$\frac{\sinh z}{z}=\prod_{n\geq 1}\left(1+\frac{z^2}{n^2 \pi^2}\right)\tag{1}$$ we have: $$\prod_{n\geq 1}\left(1+\frac{1}{n^2}\right) = \frac{\sinh \pi}{\pi}\tag{2}$$ and:
$$\sum_{k\geq 2}\log\left(1+\frac{1}{k^2}\right)=\color{red}{\log\left(\frac{\sinh \pi}{2\pi}\right)}.\tag{3}$$