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Numerical evidence suggests:

$$ - \sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} =? \sum_{n=1}^\infty \frac{1}{2^n+1} \approx 0.764499780348444 $$

Couldn't find cancellation via rearrangement.

For the second series WA found closed form.

Is the equality true?

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closed as off-topic by Todd Trimble, Chris Godsil, Ricardo Andrade, Pietro Majer, Goldstern Sep 30 '13 at 16:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Todd Trimble, Chris Godsil, Ricardo Andrade, Pietro Majer, Goldstern
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What is WA? Mathematica finds a closed form for it in terms of QPolyGamma[], not sure if that really qualifies as "closed", it could just be a name for the infinite series. $\endgroup$ – Igor Rivin Sep 30 '13 at 11:36
  • $\begingroup$ In fact, reading the documentation confirms that mathematica's closed form just renames the function. $\endgroup$ – Igor Rivin Sep 30 '13 at 11:38
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    $\begingroup$ Have you tried expanding $\frac{1}{1+2^n}=\frac{2^{-n}}{1+2^{-n}}$ using the geometric power series? Seems to me you should be able to determine whether the equality holds… $\endgroup$ – Loïc Teyssier Sep 30 '13 at 11:41
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    $\begingroup$ This might be a nice and clean example of something that's hard for computer programs like Maple, Mathematica etc. There is no simple evaluation in known functions, and to show that the two single-sums are equal, one has to write them as a double sum. Needless to say, this comment will eventually look silly as software improves... $\endgroup$ – Johan Wästlund Sep 30 '13 at 12:25
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    $\begingroup$ This problem is far too elementary for MO. The forms immediately suggest expansions into geometric series. Each of the resulting double summations can quite obviously be rearranged as $\sum_{n \geq 0} (\sum_{j k = n} (-1)^{j+1}) 2^{-n}$. I have voted to close. $\endgroup$ – Todd Trimble Sep 30 '13 at 13:08
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Write each term of the right hand side as a geometric series, $1/3 = 1/2-1/4+1/8\dots$, $1/5 = 1/4-1/16+1/64-\dots$, $1/9=1/8-1/64+1/256-\dots$ etc. Now the sum of the first terms of each series is $1/2+1/4+1/8+\dots = 1$, the sum of all second terms is $-1/4-1/16-1/64-\dots = -1/3$ etc, giving the alternating sum of the left hand side.

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This is just a formal proof using idea of Loïc Teyssier that appeared in the comments. We just rewrite both sides of the equation as double sums (using geometric series) and notice that they are equal after changing the order of summation.

Since $$\frac{q}{1-q} = \sum_{k=1}^\infty q^k,$$ we have for the left hand side

\begin{align} - \sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} & = - \sum_{n=1}^\infty \frac{(-1)^n2^{-n}}{1-2^{-n}} \\ & = - \sum_{n=1}^\infty (-1)^n\sum_{k=1}^\infty (2^{-n})^k \\ & = \sum_{k,n = 1}^\infty (-1)^{n+1} 2^{-nk}. \end{align}

Similarly, for the right hand side we use $$\frac{q}{1+q} = \sum_{k=1}^\infty (-1)^{k+1} q^k $$ to get \begin{align} \sum_{n=1}^\infty \frac{1}{1+2^n} &= \sum_{n=1}^\infty \frac{2^{-n}}{1+2^{-n}}\\ & = \sum_{n=1}^\infty \sum_{k=1}^\infty (-1)^{k+1}(2^{-n})^k \\ & = \sum_{k,n=1}^\infty (-1)^{k+1} 2^{-nk}. \end{align}

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