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

Question

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.

Answer 1

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.

Answer 2

For large $n$, the term $\sin n$ becomes neglectable and you end-up with an alternating Harmonic series. (Or you can squeeze with

$$\frac1{2(n+1)}<\frac1{2n+\sin n}<\frac1{2(n-1)}.)$$

The Harmonic series is known to be convergent but not absolutely.