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.

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.
