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I have this alternating series: $$\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n+2\sin n}$$. Leibniz test and the absolute convergence didn't work. Neither did the divergence test. When showing that $a_n=\dfrac{1}{n+2\sin n}$ is decreasing (Leibniz test) I took a function, made it's derivative and arrived nowhere. Thank you for your help!

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    $\begingroup$ Write it as $$\sum_{n = 1}^\infty (-1)^n\Biggl(\frac{1}{n} + \biggl( \frac{1}{n + 2\sin n} - \frac{1}{n}\biggr)\Biggr).$$ $\endgroup$ – Daniel Fischer Nov 22 '16 at 12:33
  • $\begingroup$ @Daniel Fischer Applying Leibniz, showing that $\dfrac{1}{n+2\sin n}-\dfrac{1}{n}$ is decreasing is not very handy. $\endgroup$ – Denis Nichita Nov 22 '16 at 12:59
  • $\begingroup$ The idea is to apply Leibniz' criterion to $\frac{(-1)^{n}}{n}$, which is pretty trivial. And $\sum (-1)^n\bigl(\frac{1}{n+2\sin n} - \frac{1}{n}\bigr)$ is absolutely convergent. Which is quite easy to see. $\endgroup$ – Daniel Fischer Nov 22 '16 at 13:01
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Take an even $n$, then

$$\frac{1}{n+2\sin n} - \frac{1}{n+1 + 2 \sin(n+1)} =\frac{1 + 2 \sin(n+1) - 2\sin n}{(n+2\sin n)(n+1 + 2 \sin(n+1))}, $$ which gives you an estimation for $n\ge 3$.

$$\left| \frac{1}{n+2\sin n} - \frac{1}{n+1 + 2 \sin(n+1)}\right| \le\frac{5}{(n-2)(n-1)} $$ The right-hand side behaves like $n^{-2}$, hence the series converges.

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  • $\begingroup$ But you've shown that $b_n=|a_n-a_{n+1}|$ converges...does this mean that $a_n$ converges? I think I'm missing sth $\endgroup$ – Denis Nichita Nov 22 '16 at 12:26
  • $\begingroup$ First, I've shown that $b_k = |a_{2k}\mathbf{+}a_{2k+1}|$ converges. Since absolute convergence implies convergence, I've also shown that $c_k = a_{2k}\mathbf{+}a_{2k+1}$ converges, too. $\endgroup$ – TZakrevskiy Nov 22 '16 at 12:34
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    $\begingroup$ Both of you are dropping the $\sum$ signs. $\endgroup$ – zhw. Nov 22 '16 at 12:35
  • $\begingroup$ @TZakrevskiy I agree. And if $\sum c_k$ converges, does it imply that $\sum a_k$ also converges? $\endgroup$ – Denis Nichita Nov 22 '16 at 13:01

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