Convergence of series $\sum_{n=1}^{\infty}\frac{1}{2^{n-1}+3}$ Where does this series converges?
Using Ratio Test, I have checked that this series converges. Now I want to find the limit of this series.
$$\sum_{n=1}^{\infty}\frac{1}{2^{n-1}+3}$$ 
I would like to have some hint.
 A: There is no simple closed form for the summation but there is one
$$S=\sum_{n=1}^{\infty}\frac{1}{2^{n-1}+3}=\frac{2 \psi _2^{(0)}\left(-\frac{\log (3)+i \pi}{\log (2)}\right)+2 i \pi +\log
   (18)}{\log (64)}$$ where appears the digamma function.
Numerically $S=0.79528027349537094149502809561786001318781240072533$
You could have a quite good approximation making
$$S=\sum_{n=1}^{\infty}\frac{1}{2^{n-1}+3}=\sum_{n=1}^{k}\frac{1}{2^{n-1}+3}+\sum_{n=k+1}^{\infty}\frac{1}{2^{n-1}+3}\approx \sum_{n=1}^{k}\frac{1}{2^{n-1}+3}+\sum_{n=k+1}^{\infty}\frac{1}{2^{n-1}}$$
$$S\approx 2^{1-k}+\sum_{n=1}^{k}\frac{1}{2^{n-1}+3}$$ Using $k=10$ would give
$\frac{99630343269857}{125276421276160}\approx 0.795284 $ which is not too bad.
Edit
Making the problem more general,
$$S=\sum_{n=1}^{\infty}\frac{1}{2^{n-1}+a}=\frac{2 \psi _2^{(0)}\left(-\frac{\log (-a)}{\log (2)}\right)+2 \log (-a)+\log
   (2)}{a \log (4)}$$
A: The limit can also be written as
$$ \frac{1}{4} + \frac{1}{5} + \frac{1}{2} \sum_{k=0}^\infty \frac{(-3/2)^k}{2^{k+1}-1}$$
But AFAIK there is no closed form.
