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!
2026-03-26 14:16:06.1774534566
Alternating series with sin
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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.