Here is another infinite sum I need you help with: $$\sum_{n=1}^\infty\frac{1}{2^n\left(1+\sqrt[2^n]{2}\right)}.$$ I was told it could be represented in terms of elementary functions and integers.

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    $\begingroup$ Where do you find all these problems? And what have you tried yourself? $\endgroup$ – Start wearing purple May 13 '13 at 22:22
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    $\begingroup$ See my comment at math.stackexchange.com/questions/389991/… And there are many problems I managed to solve myself, but I do not post them as questions at Math.SE $\endgroup$ – Laila Podlesny May 13 '13 at 22:24
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    $\begingroup$ It probably is $-1+(\ln(2))^{-1} = 0.442695\dots$ but I'll leave it to the others for a rigorous proof. $\endgroup$ – Tito Piezas III May 13 '13 at 22:27
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    $\begingroup$ Alright, have you ever read this thing? en.wikipedia.org/wiki/Layla_and_Majnun $\endgroup$ – Will Jagy May 13 '13 at 22:29
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    $\begingroup$ You remind me of Chris's sister and pals, somehow (that's positive). This being said, I'll wait for one of the local Ramanujans to find the answer. Oh, wait, someone already found something... $\endgroup$ – Julien May 13 '13 at 22:33

Note that

$$\frac{2^{-n}}{2^{2^{-n}}-1}-\frac{2^{-(n-1)}}{2^{2^{-(n-1)}}-1} = \frac{2^{-n}}{2^{2^{-n}}+1} $$

Thus we have a telescoping sum. However, note that

$$\lim_{n \to \infty} \frac{2^{-n}}{2^{2^{-n}}-1} = \frac{1}{\log{2}}$$

Therefore the sum is

$$a_1-a_0 + a_2-a_1 + a_3-a_2 + \ldots + \frac{1}{\log{2}} = \frac{1}{\log{2}}- a_0$$


$$a_n = \frac{1}{2^n \left ( 2^{2^{-n}}-1\right)}$$


$$\sum_{n=1}^{\infty} \frac{1}{2^n \left ( 1+ \sqrt[2^n]{2}\right)}= \frac{1}{\log{2}}-1$$

  • $\begingroup$ Do you have a manufacture for these ? :-) (or at least a table...) $\endgroup$ – Raymond Manzoni May 13 '13 at 22:44
  • $\begingroup$ @RaymondManzoni: I just guessed that there was an anti-difference. It was just a metter of expressing the summand in the right way to see how to go about it. $\endgroup$ – Ron Gordon May 13 '13 at 22:48
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    $\begingroup$ @RaymondManzoni: I have been working at math-contest-type problems for the better part of 25 years. I have seen a lot of tricks in that time. ;-) $\endgroup$ – Ron Gordon May 13 '13 at 22:55
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    $\begingroup$ Nicely done, Rongordonujan. $\endgroup$ – Julien May 13 '13 at 22:58
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    $\begingroup$ @user17762: one guy compares me with Ramanujan, another calls me a rat. Easy come, easy go. $\endgroup$ – Ron Gordon May 13 '13 at 23:11

Notice that $$ \frac1{2^n(\sqrt[2^n]{2}-1)} -\frac1{2^n(\sqrt[2^n]{2}+1)} =\frac1{2^{n-1}(\sqrt[2^{n-1}]{2}-1)} $$ We can rearrange this to $$ \left(\frac1{2^n(\sqrt[2^n]{2}-1)}-1\right) =\frac1{2^n(\sqrt[2^n]{2}+1)} +\left(\frac1{2^{n-1}(\sqrt[2^{n-1}]{2}-1)}-1\right) $$ and for $n=1$, $$ \frac1{2^{n-1}(\sqrt[2^{n-1}]{2}-1)}-1=0 $$ therefore, the partial sum is $$ \sum_{n=1}^m\frac1{2^n(\sqrt[2^n]{2}+1)} =\frac1{2^m(\sqrt[2^m]{2}-1)}-1 $$ Taking the limit as $m\to\infty$, we get $$ \sum_{n=1}^\infty\frac1{2^n(\sqrt[2^n]{2}+1)} =\frac1{\log(2)}-1 $$

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    $\begingroup$ Darn! I thought I would get this first because $\lim\limits_{n\to\infty}n(x^{1/n}-1)=\log(x)$ is one of my recent favorite limits, but Ron got it first. :-( $\endgroup$ – robjohn May 13 '13 at 23:03

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