Convergence of $\sum \limits ^{\infty} _{k=1} \ln(1+\frac{1}{k^2})$ I've been working on my homework, which is to discuss the convergence of $\sum \limits ^{\infty} _{k=1} \ln(1+\frac{1}{k^2})$ , but I couldn't find a solution. We were given that it could be solved by finding a majoring/minoring sequence. Up to now I know that the sequence itself converges to 0.
 A: Outline: you have $$0 \leq \ln(1+x) \leq x$$ for all $x\geq 0$ (shown e.g. by concavity of the function $f$ given by $f(x) = \ln(1+x)$).${}^{(\dagger)}$ Then use comparison theorems, noting that the series $\sum_{k=1}^\infty \frac{1}{k^2}$ converges.

${(\dagger)}$ By concavity, the continuously differentiable function $f$ stays below any of its tangents. But the tangent at $0$ is given by $g(x) = f(0)+f^\prime(0)x = 0+1\cdot x = x$.
A: The series $$\sum \limits ^{\infty} _{k=0} \ln(1+\frac{1}{k^2})$$ converges by comparison test with $$\sum _{k\ge 1}\frac{1}{k^2}$$
A: We use direct comparison of series with the p-series $\sum_{k = 0}^{\infty}\frac{1}{k^{2}}.$ Notice that for all $k$ in the given range, we have that $\ln\left(1 + \frac{1}{k^{2}}\right) < \frac{1}{k^{2}}.$ Then we have that
$$\sum_{k = 0}^{\infty}\ln\left(1 + \frac{1}{k^{2}}\right) < \sum_{k = 0}^{\infty}\frac{1}{k^{2}}.$$
Since the second comparison series converges, the given sequence must also converge.
A: $$\ln\left(1+\frac{1}{k^2}\right)=\frac{1}{k^2}+o\left(\frac{1}{k^2}\right).$$
In others words, $$\ln\left(1+\frac{1}{k^2}\right)\sim \frac{1}{k^2}.$$
The convergence follow.
A: I know I'm not really answering the actual question, which others have already answered adequately. However, I though it might still be of interest to know that the sum
$$
\sum_{k=1}^\infty \ln\left(1+\frac{1}{k^2}\right)
= \ln\left(\frac{\sinh{\pi}}{\pi}\right)
$$
which is related to the relation
$$
\sin(z\pi) = z\pi\cdot \prod_{k=1}^\infty \left(1-\frac{z^2}{k^2}\right)
$$
by entering $z=i$ making $z^2=-1$.
