Convergence of $\sum_{n=1}^{\infty} \sin\left(\frac{\pi}{2}n\right)\left(n(-1)^n+\left(\sin\frac{1}{n}\right)^{-1}\right)$

Study the convergence of the following series (this is the correct one): $$\sum_{n=1}^{\infty} \sin\left(\frac{\pi}{2}n\right)\left(n(-1)^n+\left(\sin\frac{1}{n}\right)^{-1}\right)$$

Well $\sin\left(\frac{\pi}{2}n\right)$ is an oscillating series and the use of linearity fails. My rough guess is that the series diverges.

Any tips or suggestions?

• Does the summand converge to 0? – user159517 Aug 7 '18 at 10:22
• @F.inc I have updated my answer. Be sure to check it. – xbh Aug 7 '18 at 11:49

First simplify the series. When $n$ is even, the term is $0$. For $n = 2k+1$, $\sin (\pi n /2) = \sin(\pi/2 + k\pi) = (-1)^k, k \in \mathbb N$. Hence the series becomes $$\sum_0^\infty (-1)^k \left(-(2k+1) +\left( \sin \left( \frac 1 {2k+1}\right)\right) ^{-1} \right).$$ Now consider the estimate of the term in the parentheses. Since $1/ (2k+1) \to 0 [k \to \infty]$, we have \begin{align*} -(2k+1) + \sin \left( \left(\frac 1 {2k+1}\right)\right) ^{-1} &= -(2k+1) + \left( \frac 1 {2k+1} - \frac 1{6(2k+1)^3} +O \left( \frac 1{(2k+1)^5} \right) \right)^{-1} \\ &= -(2k+1) + (2k+1) \left( 1 - \frac 1 {6 (2k+1)^2} + O\left( \frac 1 {(2k+1)^4} \right) \right)^{-1} \\ &= -(2k+1) + (2k+1) \left(1 + \frac 1{6 (2k+1)^2} + O \left( \frac 1{(2k+1)^4}\right) \right) \quad [(1 + x)^a \sim 1+ax [x \to 0]]\\ &= \frac 1 {6 (2k+1)} + O \left( \frac 1 {(2k+1)^3} \right). \end{align*}
Hence the series is $$\sum_0^\infty (-1)^k \frac 1 {6(2k+1)} + \sum_0^\infty (-1)^k O \left( \frac 1 {(2k+1)^3}\right),$$ whose first part is convergent, second part is absolutely convergent, hence the original series converges.
Denote the series by $\sum (-1)^k a_k$, this is an alternating series. The analysis above shows that for sufficiently large $k$, $$a_k - a_{k+1} = \frac {2(4k+4)}{(2k+1)^2 (2k+3)^2} + O \left( \frac 1 {k^4}\right) = \frac {8k} {(2k+1)^2(2k+3)^2} + O \left( \frac 1 {k^4}\right) \geqslant 0,$$ also $a_k \to 0$ as proved above, hence the series is a Leibniz series, and then it converges.
• A stupid question: I have not got why $-(2k+1) + (2k+1) \left( 1 - \frac 1 {6 (2k+1)^2} + O\left( \frac 1 {(2k+1)^4} \right) \right)^{-1} = -(2k+1) + (2k+1) \left( 1+ \frac 1 {6 (2k+1)^2} + O\left( \frac 1 {(2k+1)^4} \right) \right)$ – Arcticmonkey Aug 7 '18 at 12:48