# Alternating series with sin

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!

• Write it as $$\sum_{n = 1}^\infty (-1)^n\Biggl(\frac{1}{n} + \biggl( \frac{1}{n + 2\sin n} - \frac{1}{n}\biggr)\Biggr).$$ – Daniel Fischer Nov 22 '16 at 12:33
• @Daniel Fischer Applying Leibniz, showing that $\dfrac{1}{n+2\sin n}-\dfrac{1}{n}$ is decreasing is not very handy. – Denis Nichita Nov 22 '16 at 12:59
• 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. – Daniel Fischer Nov 22 '16 at 13:01

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.
• 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 – Denis Nichita Nov 22 '16 at 12:26
• 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. – TZakrevskiy Nov 22 '16 at 12:34
• Both of you are dropping the $\sum$ signs. – zhw. Nov 22 '16 at 12:35
• @TZakrevskiy I agree. And if $\sum c_k$ converges, does it imply that $\sum a_k$ also converges? – Denis Nichita Nov 22 '16 at 13:01