# If $\sum_{n=1}^\infty a_n$ is convergent, then $\sum_{n=1}^{\infty} a_n\sin(n)$

If the series $\sum_{n=1}^\infty a_n$ is convergent (absolutely convergent or conditionally convergent), then $$\sum_{n=1}^\infty a_n \sin(n)$$ is also convergent.

Any hint? today I'm not at my best. Thanks in advance.

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Have you tried to think about $(-1)^n\cdot\frac{1}{n}\to 0$? This could be some motivation. –  Christian Ivicevic Jan 7 '13 at 18:00
Could you please add a little more context? Is this a true or false question? Do you have any thoughts on whether or not it is true? –  Jonas Meyer Jan 7 '13 at 18:00
I'm sorry. Right, I actually don't now if this is true or not. –  V. Galerkin Jan 7 '13 at 18:02
Of course, absolutely convergent $\implies$ absolutely convergent. But convergent does not imply convergent (try $a_n=\sin(n)/n$). –  Did Jan 7 '13 at 18:09
@did: I think your comment should be an answer. –  sdcvvc Jan 7 '13 at 18:14

When the series is absouletely convergent, the proof is fairly straightforward as we have $$\sum_{n=1}^\infty |a_n| \geq \sum_{n=1}^\infty |a_n \sin(n)|$$

as $|\sin(n) | \leq 1$ for all n. For conditionally convergent series, I'm not too sure on how it can be proven. Indeed, for a lot of conditionally convergent series, doing anything to the $a_n$ terms can cause the new series to not converge.

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Thanks! that's actually my problem. When the series converges conditionally, is there any example of $a_n$ that makes the series $\sum a_n \sin(n)$ diverge? –  V. Galerkin Jan 7 '13 at 18:18
@V. Galerkin: Did you see did's comment? –  Jonas Meyer Jan 7 '13 at 18:20
By the looks of it, did's suggestion (in the comments of your original post) of $a_n = \frac{\sin(n)}{n}$ is such an example. –  Andrew D Jan 7 '13 at 18:21

Hint: Using the alternating series test we know that for each monotonically decreasing sequence $a_n$ with nonnegative $a_n$ and $a_n\to 0$ the series $\sum_{n=0}^{\infty}(-1)^na_n$ converges.

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Don't see why this is relevant. –  GEdgar Jan 7 '13 at 18:19
@GEdgar: I thought about that fact that $|\sin(n)|\leq 1$ like Andrew mentioned this could be an "easier" example to simplify $\sin$ if you understand. –  Christian Ivicevic Jan 7 '13 at 18:20
I think you meant an analogy: $\frac{(-1)^n}{n}$ converges, while $\frac{(-1)^n}{n} \cdot (-1)^n$ does not. –  sdcvvc Jan 7 '13 at 18:30