# On the convergence of $\sum_{n=1}^{+\infty}\frac{1}{n}\,\cos\left(\frac{\pi n}{2}\right)$

Does the following series absolutely converge, conditionally converge or diverge?

$$\sum_{n=1}^{+\infty}\frac{1}{n}\,\cos\left(\frac{\pi n}{2}\right)$$

My answer: $$0<\frac{1}{\sqrt{n}}<\left|\frac{\cos(\pi n/2)}{n}\right|$$ and $\sum_{n\geq 1}\frac{1}{\sqrt{n}}$ diverges by p-series test so by comparison test, the original series must also diverge.

• Sorry, no that is not the series i want. I'm still getting used to writing in this format. – Saeed Mohanna Oct 18 '17 at 9:31
• I want the n to be 1/n and all that multiplied by the cosine – Saeed Mohanna Oct 18 '17 at 9:32
• Yes, perfect! Thank you,! – Saeed Mohanna Oct 18 '17 at 9:37
• Do u think my answer is right? – Saeed Mohanna Oct 18 '17 at 9:38
• Your series is conditionally convergent by Dirichlet's test. The series $\sum_{n\geq 1}\frac{|\cos(\pi n/2)|}{n}$ is divergent, of course, but your conclusion is incorrect. – Jack D'Aurizio Oct 18 '17 at 9:40

This series is conditionally convergent. Its terms are: $$\frac{1}{n}\cos\left(\frac{n\pi}{2}\right) = \left\{ \begin{array}{llll} \frac{1}{n} & \mbox{if } n=4k , k>0\\ 0 & \mbox{if } n=4k+1 , k\ge0\\ -\frac{1}{n} & \mbox{if } n=4k+2 , k\ge0\\ 0& \mbox{if } n=4k+3, k\ge0\\ \end{array} \right.$$ By removing zero expression, and just considering $$n=4k$$ and $$n=4k+2$$, we could replace and change indices. By write some sentences for this series: $$\sum_{n=1}^{\infty}\frac{1}{n}\cos\left(\frac{n\pi}{2}\right)=0-\frac{1}{2}+0+\frac{1}{4}+0-\frac{1}{6}+0+\frac{1}{8}+0+...=\\-\frac{1}{2}\Big(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...\Big)=-\frac{\ln{2}}{2}\approx-0.3466$$

It's conditionally convergences because $$\sum \left|\frac{\cos\left(\frac{n\pi}{2}\right)}{n}\right|=\frac{1}{2}\Big(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\Big)$$ doesn't converge.

• What's the problem? Please write it! I think it's correct. – BarzanHayati Feb 10 at 20:36
• I could use different index for each series. – BarzanHayati Feb 10 at 20:37
• We have 4 condition for $n$ in this equation. So I could remove $\cos$ and replace it with "$0,+1,-1$". Don't write $0$ , just write $+1,-1$ – BarzanHayati Feb 10 at 20:41
• Please notice to the first equation in my solution – BarzanHayati Feb 10 at 20:42
• I corrected it! Thanks for your attention. But please don't Threaten someones for decreasing credit in such sites. I consume my time to answer question freely just like You. Users could make another account and continue to their activities. – BarzanHayati Feb 10 at 21:17

Let $$M$$ any natural number, then $$2\sum^{M}_{n=1}\cos(n\pi/2)=-1+\cos(M\pi/2)+\sin(M\pi/2).$$ Hence the partial sums of $$\cos(n\pi/2)$$ are bounded. Also $$\frac{1}{n}$$ is a null monotonic sequence. Hence from Dirichlet's test (see [1] pg. 315), the series $$\sum^{\infty}_{n=1}\frac{\cos(n\pi/2)}{n}$$ converges to a number. Moreover the series $$\sum^{\infty}_{n=1}\frac{\cos(n x)}{n^a}\textrm{, }a>0$$ converges in every interval of the form $$\epsilon\leq x\leq 2\pi-\epsilon$$, where $$\epsilon$$ any number in $$(0,\pi)$$ (see [1] pg.349).

References.

[1]: Konrad Knopp. ''Theory and application of infinite series''. Dover. (1990)