I was trying to manipulate the sum $\sum_{n=0}^\infty \frac{1}{2^n+1}=\sum_{n=0}^\infty \sum_{k=1}^\infty (-1)^{k-1}2^{-nk}$, but when we flip the summation sign, in Wolfram Alpha it says that doesn't converges to some value. In fact, if we flip, we will reach in the statement that $\sum_{n=0}^\infty \frac{1}{2^n+1}=\sum_{k=1}^\infty \frac{(-1)^{k-1}2^k}{2^k-1}=\sum_{k=1}^\infty (-1)^{k-1}\left[1+\frac{1}{2^k-1} \right]$.
The sum $\sum_{k=1}^\infty\frac{(-1)^{k-1}}{2^k-1}$ converges, but $\sum_{k=1}^\infty (-1)^{k-1}$ diverges, but this sum is known to "result" in $1/2$. If we use this value, and subtract $\sum_{k=1}^\infty\frac{(-1)^{k-1}}{2^k-1}$ on both sides, we will have that $\sum_{i=0}^\infty \left[ \frac{1}{2^i+1}-\frac{(-1)^i}{2^{i+1}-1}\right]=\frac{1}{2}$.
So, can we prove that, if we have the series $S_n=\sum_{i=0}^n \left[ \frac{1}{2^i+1}-\frac{(-1)^i}{2^{i+1}-1}\right]$, it approaches $1/2$ as n approaches infinity?
Here is the limit of $S_n$ as $n\rightarrow \infty$.
Best Answer
As noticed in the comments, we can't conclude using that $\sum_{k=1}^\infty (-1)^{k-1}=\frac12$ since it is true only as a Cesaro mean.
We can proceed, following your idea, using geometric series and flipping indices, as follows
$$\sum_{i=1}^\infty \frac{1}{2^i+1}=\sum_{i=1}^\infty \frac{2^{-i}}{1+2^{-i}}=-\sum_{i=1}^\infty\sum_{k=1}^\infty(-1)^k(2^{-i})^k=-\sum_{k=1}^\infty\frac{(-1)^k}{2^k-1}$$
and then
$$S_n=\sum_{i=0}^n \left[ \frac{1}{2^i+1}-\frac{(-1)^i}{2^{i+1}-1}\right]=$$
$$=\frac12- \frac{(-1)^n}{2^{n+1}-1}+\sum_{i=1}^n \left[ \frac{1}{2^i+1}+\frac{(-1)^{i}}{2^{i}-1}\right]\to\frac12-0+0=\frac12$$