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.

  • 1
    $\begingroup$ Please use MathJax to typeset mathematics. Here’s a tutorial. $\endgroup$ Apr 23, 2021 at 2:55
  • 1
    $\begingroup$ An edit is in the review queue. $\endgroup$
    – vb628
    Apr 23, 2021 at 3:08
  • $\begingroup$ Thank you for the edit $\endgroup$
    – Zain
    Apr 23, 2021 at 11:53
  • $\begingroup$ Why $b_{n+1} < b_n$? xpaul explains in his answer. $\endgroup$
    – GEdgar
    Apr 23, 2021 at 16:13

2 Answers 2


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.

  • $\begingroup$ To test for absolute convergence or conditional convergence, what would we do? Ratio and Root tests don't work in this case. $\endgroup$
    – Zain
    Apr 23, 2021 at 19:07
  • $\begingroup$ It is better to use ratio comparison test. $\endgroup$
    – xpaul
    Apr 23, 2021 at 19:20

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .