# Is the series $\sum_{n=1}^{\infty} \frac{(-1)^n}{2n+\sin(n)}$ absolutely convergent, convergent, or divergent?

Is this series absolutely convergent, convergent, or divergent?$$\sum_{n=1}^{\infty} \frac{(-1)^n}{2n+\sin(n)}$$

How would we show this is convergent? Alternating test? Limit comparison test?

First I did $$\lim_{n\to \infty} \frac{1}{2n+\sin(n)}$$ which is equal to zero and $$b_{n+1} < b_{n}$$. SO we know it is convergent.

Next to test absolute convergence, I did limit comparison test. $$\sum_{n=1}^\infty \bigg|\frac{(-1)^n}{2n+\sin(n)}\bigg| = \sum_{n=1}^\infty \frac{1}{2n+\sin(n)}\text{ .}$$

$$\lim_{n\to\infty} \frac{1/(2n+\sin(n))}{1/n} = 1/2$$

which is greater than zero and since $$1/n$$ diverges, then so does $$1/(2n+\sin(n))$$. Therefore $$\sum_{n=1}^\infty \frac{(-1)^n}{2n+\sin(n)}$$ is conditionally convergent.

• Please use MathJax to typeset mathematics. Here’s a tutorial. Apr 23, 2021 at 2:55
• An edit is in the review queue. Apr 23, 2021 at 3:08
• Thank you for the edit
– Zain
Apr 23, 2021 at 11:53
• Why $b_{n+1} < b_n$? xpaul explains in his answer. Apr 23, 2021 at 16:13

Let $$a_n=\frac{1}{2n+\sin(n)}.$$ Clearly $$a_n>0$$ and $$\lim_{n\to\infty}a_n=0$$. Since $$2(n+1)+\sin(n+1)-(2n+\sin(n))=2+(\sin(n+1)-\sin(n))\ge0,$$ one has $$2(n+1)+\sin(n+1)\ge2n+\sin(n)$$ or $$a_{n+1}=\frac1{2(n+1)+\sin(n+1)}\le\frac1{2n+\sin(n)}=a_n.$$ So $$\{a_n\}$$ is decreasing. By the AST, $$\sum_{n=1}(-1)^na_n$$ converges. Since $$a_n=\frac{1}{2n+\sin(n)} \sim\frac{1}{2n}$$ and $$\sum\frac{1}{2n}$$ diverges, one concludes $$\sum_{n=1}(-1)^na_n$$ converges conditionally.
For large $$n$$, the term $$\sin n$$ becomes neglectable and you end-up with an alternating Harmonic series. (Or you can squeeze with
$$\frac1{2(n+1)}<\frac1{2n+\sin n}<\frac1{2(n-1)}.)$$