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?
Best Answer
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.
Update: actually the last part of the analysis above is not right, since for series of general terms we cannot use the comparison theorem. So the right solution should be like this:
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.