# Does this series converge or diverge?

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

Thanks.

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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 $$\sum_{n=1}^{+\infty}\ln\left(1+\frac{1}{n^2}\right)\leq \sum_{n=1}^{+\infty} \frac{1}{n^2}=\frac{\pi^2}{6}.$$

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Oh, dear! Use LaTeX for your mathematics in this site. In the FAQ section you can find diretions on this... –  DonAntonio Mar 22 '13 at 0:18
I hope you don't mind, I edited your answer to make it readable. Nice elementary approach, +1. –  julien Mar 22 '13 at 0:32

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.

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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.
@user63602: Yes, that's a very correct approach. If you know about Taylor polynomials then you know that $\ln(1+x)=x+o(x^2)$ for $|x|<1$ (and for $n\to\infty$, $\frac1{n^2}<1$). –  Asaf Karagila Mar 22 '13 at 0:27