$- \sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} =? \sum_{n=1}^\infty \frac{1}{2^n+1}$ [closed]

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?

closed as off-topic by Todd Trimble♦, Chris Godsil, Ricardo Andrade, Pietro Majer, GoldsternSep 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.

• 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. – Igor Rivin Sep 30 '13 at 11:36
• In fact, reading the documentation confirms that mathematica's closed form just renames the function. – Igor Rivin Sep 30 '13 at 11:38
• 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… – Loïc Teyssier Sep 30 '13 at 11:41
• 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... – Johan Wästlund Sep 30 '13 at 12:25
• 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. – Todd Trimble Sep 30 '13 at 13:08

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