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Having that $\sum_{n=1}^{\infty}a_n$ is a convergent series and we know that $a_n > 0$ can we say that $\sum_{n=1}^{\infty}a_n \sin(a_n)$ also converges?

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closed as off-topic by Saad, Ben, user21820, user 170039, Greg Martin Dec 18 '18 at 19:20

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    $\begingroup$ What did you try? $\endgroup$ – José Carlos Santos Dec 18 '18 at 12:56
  • $\begingroup$ @JoséCarlosSantos I'm not sure if I can use something like comparision for the series that $-a_n<a_n \sin(a_n) < a_n$ but I tried only this method $\endgroup$ – avan1235 Dec 18 '18 at 12:57
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Yes, it converges, by the comparaison test and because$$(\forall n\in\mathbb{N}):\bigl\lvert a_n\sin(a_n)\bigr\rvert\leqslant\lvert a_n\rvert=a_n.$$

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    $\begingroup$ +1..Forgot about this. $\endgroup$ – xbh Dec 18 '18 at 13:02
  • $\begingroup$ Note that, since $\sum a_n$ is convergent, $\lim_{n\to\infty}a_n=0$. Since $a_n>0$, this implies that $a_n\in(0,\frac{\pi}{2})$ for $n$ large enough, so $\sin(a_n)>0$ for $n$ large enough. Therefore, you don't need the absolute value. (Of course, it makes the proof longer...) $\endgroup$ – Taladris Dec 18 '18 at 15:09
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Since $\sum a_n$ converges, for each $\varepsilon >0$, there exists $N \in \mathbb N^*$ s.t. whenever $\mathbb N^* \ni n > N, p \in \mathbb N^*$, $\sum_{n+1}^{n+p} a_k < \varepsilon$. Then $$ \left\vert \sum_{n+1}^{n+p} a_k \sin (a_k) \right\vert \leqslant \sum_{n+1}^{n+p} \vert a_k \sin (a_k ) \vert \leqslant \sum_{n+1}^{n+p} a_k < \varepsilon, $$ hence $\sum a_n \sin(a_n)$ converges by Cauchy convergence principle.

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