<s>[INST] Does $\sum_{n=1}^\infty \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)}$ have a closed form expression? It's clear that the series converges by comparison with $\sum_{n=1}^\infty \frac{1}{2^{n+1}}$, but beyond that I am not sure what can be said.
 [/INST] The first step I would suggest is to split the sum into two:
\begin{equation}
\frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} = \frac{2^n-(2^{n-1}+1)}{2^{n+1}(2^{n-1}+1)} = \frac{1}{2}\frac{1}{2^{n-1}+1}-\frac{1}{2^{n+1}}.
\end{equation}
Thus the initial sum is simply equal to
\begin{equation}
S = \frac{1}{2}\underbrace{\sum\limits_{n=0}^{\infty}\frac{1}{2^n+1}}_{S_1}-
\underbrace{\sum\limits_{n=1}^{\infty} \frac{1}{2^{n+1}}}_{S_2}.
\end{equation}
Obviously, $S_2 = \frac{1}{2}$. For the first one, WolframAlpha gives
\begin{equation}
S_1 = \sum\limits_{n=0}^{\infty}\frac{1}{2^n+1} = -1 +\frac{1}{\log 2}\psi_{1/2}^{(0)}\left(\frac{-i\pi}{\log(2)}\right).
\end{equation}
Therefore,
\begin{equation}
S = \frac{1}{2\log 2} \psi_{1/2}^{(0)}\left(\frac{-i\pi}{\log(2)}\right)-1.
\end{equation}
</s>