# Examine for absolute and conditional convergence

Examine the following series for absolute/conditional convergence: $$S_n = \sum_{n=1}^{\infty}{n^2\sin(n\pi/2)\sin(\pi/n^3)}$$

Attempt: I see that $$\sin(k\pi/2)= 0 ,\quad k=2n$$ and $$\sin(k\pi/2)=(-1)^k, \quad k=2n-1$$ In other words, the series terms are zero for all $n$ even, and when $n$ is odd, the terms oscillate in signs. So I was able to reduce the series to the following: $$S_n = \sum_{k=1}^{\infty}{(-1)^k\cdot (2k-1)^2\cdot \sin(\pi/(2k-1)^3)}$$

Using the alternating series test, I see that it satisfies the condition that $$\lim_{n \to \infty}{b_n} = 0$$ But I am unsure of how to proceed and show that the terms $b_n$ are monotonically decreasing. Also I claim that the series is not convergent absolutely,though unclear on how to show it( I was thinking of doing limit comparison test) .

• There is no $x$ on the r.h.s. in the $S_n(x)$ definition...;) – Avitus Jul 22 '13 at 21:03
• Oops!!, Edited!! – AAP Jul 22 '13 at 21:04
• So your problem is about numerical series: got it, now! – Avitus Jul 22 '13 at 21:04
• What do you know about the behaviour of $\frac{\sin x}{x}$ near $0$? – Daniel Fischer Jul 22 '13 at 21:09
• AAh!!! Why didnt I see that before, stupid of me!!! – AAP Jul 22 '13 at 21:11

## 2 Answers

As you observe, this series is equivalent to $$\sum\limits_{n \text{ odd}}(-1)^{(n+1)/2}n^2\sin(\pi/n^3).$$ Using the series expansion $\sin x = x - \frac{1}{3!} x^3 + \frac{1}{5!}x^5-\cdots$ we get that $$\sum\limits_{n \text{ odd}}(-1)^{(n+1)/2}n^2\sin(\pi/n^3)=\sum\limits_{n \text{ odd}}(-1)^{(n+1)/2}n^2\left(\frac{\pi}{n^3}-\frac{\pi^3}{3!\cdot n^9}+\frac{\pi^5}{5!\cdot n^{15}}-\cdots\right)$$ and observing that $$\left|(-1)^{(n+1)/2}n^2\left(-\frac{\pi^3}{3!\cdot n^9}+\frac{\pi^5}{5!\cdot n^{15}}-\cdots\right)\right|\leq \frac{\pi^3}{3!\cdot n^7}$$ we see that the series $$\sum\limits_{n \text{ odd}}(-1)^{(n+1)/2}n^2\left(-\frac{\pi^3}{3!\cdot n^9}+\frac{\pi^5}{5!\cdot n^{15}}-\cdots\right)$$ is absolutely convergent, so the absolute/conditional convergence of our original series is determined by the series $$\sum\limits_{n \text{ odd}}(-1)^{(n+1)/2}n^2\frac{\pi}{n^3}=\pi\sum\limits_{n\text{ odd}}(-1)^{(n+1)/2}\frac{1}{n}$$ which is easily seen to be conditionally but not absolutely convergent.

You just find $$S = \sum_{k=1}^{\infty}{(-1)^k\cdot (2k-1)^2\cdot \sin(\pi/(2k-1)^3)}=\sum_{k=1}^{\infty}u_k$$ so to complete we use the Taylor series: $$\sin(\pi/(2k-1)^3)=\frac{\pi}{(2k-1)^3}+O\left(\frac{1}{(2k-1)^4}\right)$$ hence $$u_k=\underbrace{\frac{\pi(-1)^k}{(2k-1)}}_{v_k}+\underbrace{O\left(\frac{1}{(2k-1)^2}\right)}_{w_k}$$ so the given series is convergent since it's a sum of two convergent series but it isn't absolutely convergent since $\displaystyle\sum_k |v_k|$ is divergent.