Does the series $\sum\limits_{n=1}^{ \infty}n\tan\left( \dfrac { \pi}{2^{n+1}}\right)$ converge or diverge? My idea was to use the limit comparison test and $\sum\limits_{n=1}^{\infty} \dfrac {n}{2^{n}}$, but then I don't know what to do with the tangent which in the limit is 0.

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    $\begingroup$ Please consider using \tan to get $\tan$ instead of $tan$. For example, your $ntan$ would appear as $n\tan$. Also, consider using \left( and \right) to get larger brackets, e.g. $\left(\frac{1}{2}\right)$ instead of $(\frac{1}{2})$. $\endgroup$ – Fly by Night Mar 21 '14 at 21:10
  • $\begingroup$ @Cameron: Is there a particular reason for the rollback? $\endgroup$ – Asaf Karagila Mar 21 '14 at 21:15
  • $\begingroup$ I assume your index should be $n$, not $i$. $\endgroup$ – Andrew Kelley Mar 21 '14 at 21:17
  • $\begingroup$ @AsafKaragila Only just because of dfrac. You edited it immediately after I finished editing but I wanted to keep the fraction relatively readable in the inline equation. $\endgroup$ – Cameron Williams Mar 21 '14 at 21:47
  • $\begingroup$ @Dror Check out the edits to the question $\endgroup$ – Nick Mar 21 '14 at 21:58

Hint: note that $$ \lim_{n \to \infty} \frac{\tan\left(\frac{\pi}{2^{n+1}}\right)}{\left(\frac{\pi}{2^{n+1}}\right)} = 1 $$

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  • $\begingroup$ Where did the n in front of the tangent function go? $\endgroup$ – Nick Mar 21 '14 at 21:57
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    $\begingroup$ This isn't a direct manipulation of the sum; this is just a fact that you can use. It will help if you want to use the ratio test on the sum itself. $\endgroup$ – Ben Grossmann Mar 21 '14 at 22:05

We have that, as $(\tan x)'=\sec^2 x$, then, for every $x\in [0,\pi/4]$, there exists a $\vartheta\in (0,1)$, such that $$ \tan x=\tan x -\tan 0=x\sec^2 \vartheta x, $$ and hence $$ \lvert\tan x\rvert \le 2\lvert x\rvert, $$ since for $x\in [0,\pi/4]$ $$ 1\le \sec x\le \sqrt{2}. $$ Clearly, $\dfrac{\pi}{2^{n+1}}\in [0,\pi/4]$, and thus $$ 0\le \tan \left(\frac{\pi}{2^{n+1}}\right)\le \frac{\pi}{2^n}, $$ and as the series $\sum_{n=1}^\infty n2^{-n}$ converges (i.e., ratio test), so does the series $$ \sum_{n=1}^\infty n\tan \left(\frac{\pi}{2^{n+1}}\right), $$ and it does absolutely.

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  • $\begingroup$ I think you mean that $(\tan x)' = \sec^2 x$.. :) $\endgroup$ – Cameron Williams Mar 21 '14 at 21:48
  • $\begingroup$ @CameronWilliams: I corrected it. Thanx. $\endgroup$ – Yiorgos S. Smyrlis Mar 21 '14 at 21:53
  • $\begingroup$ @YiorgosS.Smyrlis. Any idea about the result of the summation ? The value $3.40841$ does not look very appealing to RIES. $\endgroup$ – Claude Leibovici Mar 22 '14 at 7:01

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