Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $\{a_n\}$ is a sequence of positive terms such that the series $$\sum_{n=1}^\infty a_n$$ coverges, does the series $$\sum_{n=1}^\infty \sin a_n$$ also converge?

I believe that limit comparison test is necessary but I'm not sure how to use it here

share|cite|improve this question
Don't use the title's body and the question's body as one, please. Type your question into the question body. – Git Gud Mar 21 '13 at 18:55
up vote 9 down vote accepted

Since $\lim\limits_{k \to \infty} a_k=0$, $$\tag 1\lim_{k \to \infty} \frac{\sin a_k}{a_k}=1$$

Thus $\sum_k |\sin a_k|$ converges $\iff \sum_k |a_k|=\sum_k a_k$ does. By your hypothesis and the above, $\sum_k \sin a_k$ will be absolutely convergent, so it will converge.

share|cite|improve this answer
In the above limit would k be approaching infinity? I'm not quite sure how the limit equals 1 – Fred Mar 21 '13 at 19:12
@Michael Recall that $$\lim_{x\to 0}\frac{\sin x}x=1$$ Yes, as $k\to\infty$; $a_k\to 0$. Since $\sin$ is continuous, $(1)$ above follows. – Pedro Tamaroff Mar 21 '13 at 19:22
Sorry, deleted my stupid comment...+1. – 1015 Mar 21 '13 at 19:23
@julien I'm curious! What did it say? =P – Pedro Tamaroff Mar 21 '13 at 19:25
It said: what about $a_k=k\pi$? I had not read the very first line of your answer. – 1015 Mar 21 '13 at 19:26

use that $|\sin(y)|\leq |y| $ (prove with the series) else you could prove it with mean value theorem.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.