Does $\sum_{n=1}^{\infty}\ln(n\sin(\frac{1}{n}))$ converge? I must determine whether the following series converges: $$\sum_{n=1}^{\infty}\ln\left(n\sin\left(\frac{1}{n}\right)\right)$$
I know that in general, I must use the limit comparison test, but I cannot find an expression to which I can compare it. For instance, I have tried the usual process:
For $n$ large, we have that $\lim_{n\to \infty}n\sin\frac1n=1$, and so, $\ln(1)=0$. This fails the divergence test, but it cannot be concluded automatically that the series is convergent either. How may I proceed here? Any help would be appreciated.
 A: 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.
A: $$S=\sum_{n\geq 1}\left(1-n\sin\frac{1}{n}\right)$$
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 $0.2652\leq S\leq 0.2654$.
A: Since $$ \sin h \sim h-\frac{h^3}6 ~~\text{and}~~~\lim_{x\to 0}\frac{\ln(x+1)}{x} = 1$$
we have, 
$$\lim_{n\to\infty} \color{red}{n^2}\ln\left(n\sin\left(\frac{1}{n}\right)\right)=\lim_{h\to 0} \frac{\ln\left(\frac{\sinh-h}{h}+1\right)}{h^2} \sim \lim_{h\to 0}\frac{1}6 \frac{\ln\left(1-\frac{h^2}6\right)}{\frac{-h^2}6} =\frac{1}6 $$
Then, the exist N such that for $n>N$ we have, 
$$\left|\ln\left(n\sin\left(\frac{1}{n}\right)\right)\right|\le \frac{c}{n^2}$$
this prove the convergence of 
$$\sum_{n=1}^{\infty}\ln\left(n\sin\left(\frac{1}{n}\right)\right)$$
since, $$\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$$
A: A variant using the equivalent of $\ln x$ near $x=1$:
As this is a series with negative terms, we can use equivalents. 
Note that $\;\lim\limits_{n\to\infty}n\sin\dfrac1n=1$, and remember that if $x\to 1$, $\:\ln x\sim x-1$. Thus, using Taylor's formula at order $3$,
$$\ln n\sin\dfrac1n \sim_{n\to\infty}\:n\sin\frac1n-1=n\Bigl(\frac1n-\frac1{6n^3}+o\Bigl(\frac1{n^3}\Bigr)\Bigr)-1=-\frac1{6n^2}+o\Bigl(\frac1{n^2}\Bigr) \sim_{n\to\infty}-\frac1{6n^2}, $$
which is a convergent series.
A: Taylor series of $\sin x$ is 
$$\sin x = x - \frac{x^3}{3!} + O(x^5)$$
Hence
$$\sin \frac{1}{n} = \frac{1}{n}- \frac{1}{6n^3} + O(\frac{1}{n^5})$$
So
$$n\sin \frac{1}{n} = 1- \frac{1}{6n^2} + O(\frac{1}{n^4})$$
Taylor series of $\ln$ is 
$$\ln(1+x) = x + O(x^2)$$
For $x = -\frac{1}{6n^2} + O(\frac{1}{n^4})$, we get
$$\ln(1 - \frac{1}{6n^2} + O(\frac{1}{n^4})) = -\frac{1}{6n^2} + O(\frac{1}{n^4})$$
Then you'd get a sum of convergent series since all those sums have $p > 1$ according to the p-series test.
A: Taylor's theorem tells us that $\sin x = x - \dfrac1{3!} x^3 + \dfrac1{5!} \cos(\xi) x^5$ for some $\zeta \in [0,x]$ whenever $x > 0$. Since $0 \le \cos \xi \le 1$ when $x = \dfrac1n$:
$$x - \frac1{3!} x^3 \le \sin x \le x - \frac1{3!} x^3 + \frac1{5!}x^5$$
Substituting $x = \dfrac1n$:
$$\frac1n - \frac1{3!} \frac1{n^3} \le \sin \frac1n \le \frac1n - \frac1{3!} \frac1{n^3} + \frac1{5!} \frac1{n^5}$$
Multiplying all sides by $n$:
$$1 - \frac1{3!} \frac1{n^2} \le n \sin \frac1n \le 1 - \frac1{3!} \frac1{n^2} + \frac1{5!} \frac1{n^4}$$

Applying Taylor's theorem on $\ln(1-x)$ when $x>0$:
$$\ln(1-x) = - \frac 1 {1 - \xi} x$$
for some $\xi \in [0,x]$
Therefore:
$$-\frac{x}{1-x} \le \ln(1-x) \le -x$$
That is:
$$1-\frac{1}{1-x} \le \ln(1-x) \le -x$$

Combining the above two sections:
$$\ln \left(1 - \frac1{3!} \frac1{n^2}\right) \le \ln \left( n \sin \frac1n \right) \le \ln \left(1 - \frac1{3!} \frac1{n^2} + \frac1{5!} \frac1{n^4} \right)$$
Then:
$$\frac{-1}{6n^2-1} \le \ln \left( n \sin \frac1n \right) \le - \frac1{3!} \frac1{n^2} + \frac1{5!} \frac1{n^4}$$
Applying sandwich theorem:
$$\sum_{n=1}^\infty \frac{-1}{6n^2-1} \le \sum_{n=1}^\infty \ln \left( n \sin \frac1n \right) \le \sum_{n=1}^\infty - \frac1{3!} \frac1{n^2} + \frac1{5!} \frac1{n^4}$$
and both sums can be easily shown convergent.
