# Study the conditional or absolute convergence of $\sum _{n=1}^{\infty }\ln\left(n\cdot \sin\left(\frac{1}{n}\right)\right)$

Here I have the following series:

$$\sum _{n=1}^{\infty }\:\ln\left(n\cdot \sin\left(\frac{1}{n}\right)\right)$$

$$\ln\left(n\cdot \sin\left(\frac{1}{n}\right)\right)$$ would be negative for any integer $$n\ge1$$, so taking the absolute value of it would let the series left to study be:

$$\sum _{n=1}^{\infty }\:-\ln\left(n\cdot \sin\left(\frac{1}{n}\right)\right)$$

To make it easier, that series according to Wolfram Mathematica would converge. All left to do is to prove it.

Now, in my opinion, the ratio and root tests don't seem to be very beneficial for this problem in particular. So at the time of choosing a test, the direct comparison and limit comparison tests are, to me, appropriate. But when I attempted using the first one, I just couldn't find a boundary that would work to prove the convergence. Kind of the same story repeated with the limit comparison test. I'm stuck. Thanks for your answer in advance.

• Hint: Apply the Limit Comparison Test with the $p$-series $\sum\frac{1}{n^2}$. In doing so, the Taylor approximation $$\log\left(\frac{\sin x}{x}\right)=-\frac{x^2}{6}+\dots$$ near $x=0$ might be helpful. Apr 24 '20 at 23:11

• $$\sin \frac1n=\frac 1n-\frac 1{6n^3}+o\bigl(\frac1{n^3}\bigr)$$, so $$n\sin\frac 1n = 1-\frac 1{6n^2}+o\Bigl(\dfrac1{n^2}\Bigr)$$
• This means that $$\:n\sin\frac 1n \sim_\infty 1-\frac 1{6n^2}$$, whence $$\ln\Bigl(n\sin\frac 1n\Bigr)\sim_\infty\ln\Bigl(1-\frac 1{6n^2}\Bigr)\sim_\infty-\frac 1{6n^2},$$ with is a convergent $$p$$-series.
• Writing sums in equivalents and taking $\ln$ of equivalents is dangerous. It is also true that $n \sin(1/n) \sim 1 +1/n$, but $\ln(n\sin(1/n))$ is not equivalent to $1/n$. Apr 24 '20 at 23:55