$ - \sum=?iso-8859-1?Q?=5F{n=3D1}^\infty_\frac{?=(-1)^n}{2^n-1} =? \sum_{n=1}^\infty \frac{1}{2^n+1}$ – mathoverflow.net

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

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 ...