# Alternating series test $\sum_{n\ge2}\ln\left(1+(-1)^n\sin\left(\frac1n\right)\right)$

My question is about one of the alternating test conditions.

Let this example to illustrate my comments

$\sum_{n\ge2} u_n$ with $u_n=\ln\left(1+(-1)^n\sin\left(\dfrac{1}{n}\right)\right)$

May I use the taylor expansion to do the test??

For instance $\sin\left(\dfrac{1}{n}\right)=\dfrac{1}{n}+o(n^{-2})$ so $u_n=(-1)^n\left(\dfrac{1}{n}+o(n^{-2})\right)$

Let $v_n=\dfrac{1}{n}$ and $w_n=o(n^{-2})$

Should I test $(v_n+w_n)$ and how to show $w_n$ is strictly decreasing for all $n$?

• "how to show $w_n$ is strictly decreasing for all $n$?" What for? One does not want to apply the alternating series theorem to $\sum w_n$... – Did Dec 28 '17 at 23:33
• @Did Ok, thanks, we apply the absolute convergence to $\sum w_n$ – Stu Dec 28 '17 at 23:36

Since

$$\sin\left(\dfrac{1}{n}\right)=\dfrac{1}{n}+o\left(\dfrac{1}{n^2}\right)$$

thus

$$u_n=(-1)^n\left(\dfrac{1}{n}+o\left(\dfrac{1}{n^2}\right)\right)=\dfrac{(-1)^n}{n}+(-1)^no\left(\dfrac{1}{n^2}\right)=v_n+w_n$$

$w_n$ is absolutely convergent.

• $\sin\left(\dfrac{1}{n}\right)=\dfrac{1}{n}+\mathcal{O}\left(\dfrac{1}{n^3}\right)$ big O is better I think – Stu Dec 29 '17 at 1:49
• @Stu ops...of course you are right! just a typo, I prefer little-o notation! – user Dec 29 '17 at 1:50

Let's have a look at the summand: $$$\lim_{n \to \infty} \ln\left(1+(-1)^n\sin(\frac{1}{n})\right)= \lim_{n \to \infty}\ln\left(1+\frac{1}{n\cdot (-1)^n}\right)= \frac{(-1)^n}{n}$$$

Using the alternate sign test in the last expression, you have the convergence