Does $\sum_{n=1}^{\infty}\ln(n\sin(\frac{1}{n}))$ converge? [duplicate]

Solution 1:

Using Taylor series, $$\sin x\sim x-\frac{x^3}{3!}+o(x^3)$$$$\ln(1+x)\sim x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)$$ When $n\rightarrow\infty$, $$n\sin\frac{1}{n}\sim n(\frac{1}{n}-\frac{1}{3!n^3}+o(\frac{1}{n^3}))=1-\frac{1}{6n^2}+o(\frac{1}{n^2})$$

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

As $\sum\dfrac{1}{n^2}$ is convergent, so is $\sum\ln(n\sin\dfrac{1}{n})$.

Hope this can help you.

Solution 2:

$$S=\sum_{n\geq 1}\left(1-n\sin\frac{1}{n}\right)$$ converges and so does your series by $0\leq \log(1+x)\leq x$ over $[0,1]$.
$S$ has a nice integral representation: the sine function is entire, hence

$$ S = \sum_{n\geq 1}\sum_{m\geq 1}\frac{(-1)^{m+1}}{(2m+1)! n^{2m}}=\sum_{m\geq 1}\frac{\zeta(2m)(-1)^{m+1}}{(2m+1)!}=2\int_{0}^{+\infty}\!\!\!\!\underbrace{\frac{\text{Ber}_2(x)}{e^{x^2/4}-1}}_{\text{gaussian-shaped}}dx $$ by the integral representation for the $\zeta$ function, where $\text{Ber}_2$ is a Kelvin function.
By the fast-convergent series representation and Leibniz' test we have $0.2652\leq S\leq 0.2654$.