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How to evaluate $\displaystyle\sum_{n=1}^\infty \dfrac{1}{2^n}\tanh{\left(\dfrac{1}{2^n}\right)}$ ?

In general, what is the explicit form of $\displaystyle\sum_{n=1}^\infty x^n\tanh{x^n}$ ?

Thanks in advance.

Bless
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    the sum is $\frac{2}{e^2-1}$ – E.H.E Nov 26 '15 at 11:14
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    Hint: For the first series, use the fact that $\tanh x=2\coth 2x-\coth x$ and a telescoping argument. By the way, the first series is very similar to this question. – mickep Nov 26 '15 at 11:21
  • @mickep i ask myself if jack's approach also works for the general case – tired Nov 26 '15 at 11:30
  • @tired I encourage you to try it out. I'm cooking, so I don't have the possibility to grab pen and paper at the moment (and when I'm done, I'm sure someone will have tried) :) – mickep Nov 26 '15 at 11:33
  • I think the question is duplicate https://math.stackexchange.com/questions/1364986/evaluating-sum-n-0-infty-2-n-tanh-2-n?rq=1 – E.H.E Nov 26 '15 at 11:36
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    @mickep ok, i think i doesn't because the special choice of $x=1/2$ yields some cancelations which makes it possible to aplly the identity (3) in jack's answer. At least some more work has to be done at this stage of the calculations which is beyond my personal (time-)horizon at the moment.. :( – tired Nov 26 '15 at 12:02

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