# Does $\sum_{n=1}^{\infty} \ln(n\cdot \sin(\frac{1}{n}))$ converges or diverges?

$$\sum_{n=1}^{\infty} \ln(n\cdot \sin\left(\frac{1}{n}\right))$$

Does it converge or diverge?

My instincts say that I should use the comparison test, but compare it with what other sequence?

Any other test does not seem helpful here.

Thanks!

• How does $n\sin \frac{1}{n}$ behave for large $n$? Hence how does $\ln \bigl(n\sin \frac{1}{n}\bigr)$ behave? – Daniel Fischer Feb 22 '16 at 16:49

1. As $n\to\infty$, $\sin\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{6n^3}+o\left(\frac{1}{n^3}\right)$

2. Thus, $\ln n\sin\left(\frac{1}{n}\right)=\ln\left(1-\frac{1}{6n^2}+o\left(\frac{1}{n^2}\right)\right)=\frac{-1}{6n^2}+o\left(\frac{1}{n^2}\right)$

3. So, the terms of our sequence are asymptotic to $\frac{-1}{6n^2}$, which has convergent sum, and thus our series also has a convergent sum by the comparison test.

• Thanks! What I still don't fully understand is the end of step 2. How did you conclude that the argument of ln equals -1/6n^2? – MorKadosh Feb 22 '16 at 17:18
• This comes from the Taylor expansion: $\log (1+x)=x+o(x)$ as $x\to0$ – πr8 Feb 22 '16 at 17:20

The Weierstrass product for the sine function gives: $$\frac{\sin z}{z}=\prod_{m\geq 1}\left(1-\frac{z^2}{\pi^2 m^2}\right)\tag{1}$$ from which: $$\log\left(\frac{\sin z}{z}\right) = \sum_{m\geq 1}\log\left(1-\frac{z^2}{\pi^2 m^2}\right) = -\sum_{m\geq 1}\sum_{k\geq 1}\frac{z^{2k}}{k\cdot \pi^{2k} m^{2k}}=-\sum_{k\geq 1}\frac{z^{2k}\zeta(2k)}{k\cdot\pi^{2k}}\tag{2}$$ then, by replacing $z$ with $\frac{1}{n}$ and summing over $n\geq 1$,

$$\color{red}{S}=\sum_{n\geq 1}\log\left(n\cdot\sin\frac{1}{n}\right) = \color{red}{-\sum_{k\geq 1}\frac{\zeta(2k)^2}{k\cdot \pi^{2k}}}\tag{3}$$

where the RHS is clearly a fast-converging series, since $\zeta(2k)=1+O\left(\frac{1}{k}\right)$.

Numerically, we have $\color{red}{S}\approx -0.280556336$.