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.
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.
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 $x &gt; 0$. 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 $x&gt;0$:
$$\ln(1-x) = - \frac 1 {1 - \xi} x$$
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.