Convergence of series $\sum_{n=1}^{\infty}\frac{1}{2^{n-1}+3}$

Question

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.

Answer 1

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,

$$S=\sum_{n=1}^{\infty}\frac{1}{2^{n-1}+a}=\frac{2 \psi _2^{(0)}\left(-\frac{\log (-a)}{\log (2)}\right)+2 \log (-a)+\log (2)}{a \log (4)}$$

Answer 2

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.