Summation $\sum_{n=1}^\infty \frac{1}{2^n+1}$

Question

So I was was messing around with infinite series and I came across one that is deceptively similar to the familiar $\sum_{n=1}^\infty \frac{1}{2^n} = 1 $ Simply add $1$ to each denominator of each term of the series. This is $\sum_{n=1}^\infty \frac{1}{2^n+1} \approx 0.76449...$ I have used all my weaponry on trying to crack it and the best I've come up with is the equivalent series $\sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} $ Could somebody help me retrieve some sort of compact value for this expression ( e.g. $\sum_{n=1}^\infty \frac{1}{n^2} $ can be rewritten as $\frac{\pi^2}{6}$) or is this pretty much impossible? Thanks!

Answer 1

There is no known closed form for this sum.

Answer 2

You could use the fact that $$\frac{1}{2^n+1}+\frac{1}{2^n(2^n+1)} = \frac{1}{2^n},$$ and then attempt to find a closed form for the following sum, which appears to be equally difficult,

$$\sum_{n=1}^\infty \frac{1}{2^n(2^n+1)}$$ and then subtract the closed form of the previous sum from $1.$