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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.

Laila Podlesny
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2 Answers2

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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

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

where

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

or

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

Ron Gordon
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  • Do you have a manufacture for these ? :-) (or at least a table...) – Raymond Manzoni May 13 '13 at 22:44
  • @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. – Ron Gordon May 13 '13 at 22:48
  • Hmmm... and in the case of the sum of $\cot{x}-2\cot{2 x}=\tan{x}$ here ? (for those who missed this nice trick) – Raymond Manzoni May 13 '13 at 22:50
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    @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. ;-) – Ron Gordon May 13 '13 at 22:55
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    Nicely done, Rongordonujan. – Julien May 13 '13 at 22:58
  • @julien: of all the compliments I have been paid (hopefully more numerous than the insults I have been paid), that is the first time I have ever been called that. Yet again, you are too kind. – Ron Gordon May 13 '13 at 23:01
  • Congratulations on getting it first (+1). Since I put some effort into my solution, I hope you don't mind if I leave it up. – robjohn May 13 '13 at 23:05
  • @robjohn: be my guest. I would be deluding myself if I thought that I was the only one on this site that could possibly answer this question. – Ron Gordon May 13 '13 at 23:07
  • @robjohn Two rats bet me :-) +1 to both the rats. Last half hour of my effort gone! –  May 13 '13 at 23:08
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    @user17762: one guy compares me with Ramanujan, another calls me a rat. Easy come, easy go. – Ron Gordon May 13 '13 at 23:11
  • @RonGordon The rats was supposed to be a complement :). –  May 14 '13 at 00:30
  • @user17762: heh, I know, just a little dark humor on my part. Thanks. – Ron Gordon May 14 '13 at 11:07
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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 $$ \sum_{n=1}^\infty\frac1{2^n(\sqrt[2^n]{2}+1)} =\frac1{\log(2)}-1 $$

robjohn
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    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. :-( – robjohn May 13 '13 at 23:03