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If $\sum_{n=1}^{\infty} a_{n}$ is convergent then I was thinking about the convergence of $\sum_{n=1}^{\infty} n a_{n} \sin\left(\frac{1}{n}\right)$.

I think that the series is convergent as $\sum_{n=1}^{\infty} n a_{n} \sin\left(\frac{1}{n}\right) \leq \sum_{n=1}^{\infty} n a_{n} \frac{1}{n} = \sum_{n=1}^{\infty} a_{n} $ so

$\sum_{n=1}^{\infty}n a_{n} \sin\left(\frac{1}{n}\right) \leq \sum_{n=1}^{\infty}a_{n}$ and hence by comparision test the series is convergent,is this correct,also any other approach to this?

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    $\begingroup$ The result itself is true, but the reason is more delicate if you do not assume positivity of $a_n$ (so that you are no longer allowed to invoke comparison test). $\endgroup$ – Sangchul Lee Sep 29 '17 at 10:54
  • $\begingroup$ If $a_n \geq 0$ it's trivial. $\endgroup$ – Rafael Gonzalez Lopez Sep 29 '17 at 10:58
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Your argument works, and is probably the most straightforward argument if the $a_n$ are assumed nonnegative, or - adding in absolute values - if $\sum a_n$ is absolutely convergent.

If $\sum a_n$ is only conditionally convergent, we must take the difference $\frac{1}{n} - \sin \bigl(\frac{1}{n}\bigr)$ into account. Since

$$0 < \frac{1}{n} - \sin \biggl(\frac{1}{n}\biggr) < \frac{1}{6n^3}$$

by Taylor expansion of the sine, writing

$$na_n\sin\biggl(\frac{1}{n}\biggr) = a_n - \underbrace{na_n\Biggl(\frac{1}{n} - \sin \biggl(\frac{1}{n}\biggr)\Biggr)}_{b_n}$$

shows the convergence of the series, as $a_n$ is bounded and hence

$$\Biggl\lvert na_n\Biggl(\frac{1}{n} - \sin \biggl(\frac{1}{n}\biggr)\Biggr)\Biggr\rvert \leqslant \frac{\lvert a_n\rvert}{6n^2}$$

gives the absolute convergence of $\sum b_n$, whence convergence of $\sum (a_n - b_n)$.

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