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

 A: 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}
A: 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.
