Convergence of $\sum\ln(n\sin\frac{1}{n})$ How to prove that $\sum\ln(n\sin\frac{1}{n})$ converges rigorously?
The idea is to expand $\sin(t)$ and $\ln(1+t)$. But I'm not sure how to do it rigorously. We have $n\sin(1/n)=n(1/n+O(1/n^3))=1+nO(1/n^3)$ as $n\to \infty$. Then I need to substitute it into $\ln (1+t)=t-t^2/2+O(t^3)$. (Or should I use the little o in one of both cases?) This involves taking powers of $nO(1/n^3)$ (or of the little o) and multiplying it by functions, and I'm not very familiar how the o's behave under such operations (and not aware of any source to find that out). So how to do the prove in a rigorous way?
 A: Yes by Taylor’s expansion we have


*

*$\sin \left(\frac1n \right)= \frac1n+O\left(\frac1{n^3}\right)$


therefore 
$$\ln\left(n\sin\frac{1}{n}\right)=\ln (1+O(1/n^2))=O\left(\frac1{n^2}\right)$$
and the given series converges by limit comparison test with $\sum 1/n^2$.
More in detail we have


*

*$\sin \left(\frac1n \right)= \frac1n-\frac1{6n^3}+O\left(\frac1{n^5}\right)$


therefore 
$$\ln\left(n\sin\frac{1}{n}\right)=\ln \left(1-\frac1{6n^2}+O\left(\frac1{n^4}\right)\right)=$$
$$=-\frac1{6n^2}+O\left(\frac1{n^4}\right)+O\left[\left(-\frac1{6n^2}+O\left(\frac1{n^4}\right)\right)^2\right]=-\frac1{6n^2}+O\left(\frac1{n^4}\right)$$
indeed
$$O\left[\left(-\frac1{6n^2}+O\left(\frac1{n^4}\right)\right)^2\right]=O\left[\frac1{36n^4}-2\frac1{6n^2}O\left(\frac1{n^4}\right)+O\left(\frac1{n^8}\right)\right]=$$
$$=O\left[\frac1{36n^4}+O\left(\frac1{n^6}\right)+O\left(\frac1{n^8}\right)\right]=O\left[\frac1{36n^4}+O\left(\frac1{n^6}\right)\right]=O\left(\frac1{n^4}\right)$$
indeed by definition
$$\frac1{36n^4}+O\left(\frac1{n^6}\right)=O\left(\frac1{n^4}\right)$$
since
$$\left|\frac{\frac1{36n^4}+O\left(\frac1{n^6}\right)}{\frac1{n^4}}\right|\to \frac1{36}$$
and
$$O\left[O\left(\frac1{n^4}\right)\right]=O\left(\frac1{n^4}\right)$$
since
$$\left|\frac{O\left(\frac1{n^4}\right)}{\frac1{n^4}}\right|\to L< \infty$$
and finally
$$\ln\left(n\sin\frac{1}{n}\right)=-\frac1{6n^2}+O\left(\frac1{n^4}\right)=O\left(\frac1{n^2}\right)$$
indeed by definition
$$\left|\frac{-\frac1{6n^2}+O\left(\frac1{n^4}\right)}{\frac1{n^2}}\right|\to \frac1{6}$$
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: The series is convergent.
For large $n$ you can write $n \sin(\frac{1}{n}) = 1-\frac{1}{6 n^2} +O(1/n^4)$ and because $\log(1-x) \simeq -x$ the summand becomes $-\frac{1}{6 n^2}$ and this gives the convergent series for the Riemann zeta function of argument 2.
