# 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.

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

## 4 Answers

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.

• +1 This is a elegant and short as one can expect...yet in some places (yup, the Hebrew University's Math department, at least back in the 80's) they use to study first infinite series and later (much later, proportionally) Taylor, power and stuff series... – DonAntonio Mar 22 '13 at 0:14
• But how would you work with Taylor series without having worked out the basics of infinite series first? In BGU we studied Taylor polynomials in calculus I, then infinite series and all the shizzle in calculus II. – Asaf Karagila Mar 22 '13 at 0:23
• hmmmm... im thinking using the Limit Comparison Test with 1/n^2. But I'm getting mixed up with the limit, so now I'm not so confident... – user63602 Mar 22 '13 at 0:26
• @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
• ahh okay i got it. I got the condition of LCT confused with that of the ratio test. Thanks! – user63602 Mar 22 '13 at 0:36

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

• 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.

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 $$\sum_{n\geq1}\log\left(1+\frac{1}{n^{2}}\right)=\log\left(\prod_{n\geq1}\left(1+\frac{1}{n^{2}}\right)\right)=\color{red}{\log\left(\frac{\sinh\left(\pi\right)}{\pi}\right)}.$$