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.

  • $\begingroup$ Sorry, no that is not the series i want. I'm still getting used to writing in this format. $\endgroup$ – Saeed Mohanna Oct 18 '17 at 9:31
  • $\begingroup$ I want the n to be 1/n and all that multiplied by the cosine $\endgroup$ – Saeed Mohanna Oct 18 '17 at 9:32
  • $\begingroup$ Yes, perfect! Thank you,! $\endgroup$ – Saeed Mohanna Oct 18 '17 at 9:37
  • $\begingroup$ Do u think my answer is right? $\endgroup$ – Saeed Mohanna Oct 18 '17 at 9:38
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.

References: https://en.wikipedia.org/wiki/Alternating_series_test

  • $\begingroup$ What's the problem? Please write it! I think it's correct. $\endgroup$ – BarzanHayati Feb 10 at 20:36
  • $\begingroup$ I could use different index for each series. $\endgroup$ – BarzanHayati Feb 10 at 20:37
  • $\begingroup$ 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 $+1,-1$ $\endgroup$ – BarzanHayati Feb 10 at 20:41
  • $\begingroup$ Please notice to the first equation in my solution $\endgroup$ – BarzanHayati Feb 10 at 20:42
  • $\begingroup$ 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. $\endgroup$ – 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).


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


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