calculus - Is it possible to give the sum of the convergent series $sum_{j=1}^{infty}frac{1}{1+2^j}?$

It is easy to show that the series $\sum_{j=1}^{\infty}\frac{1}{1+2^j}$ is convergent. Now my curiosity is, Is it possible to give the sum of this series? I have tried by Mathematica to evaluate it, and the output is:

In[1]:= Sum[1/(1 + 2^j), {j, 1, Infinity}]

Out[1]= (-Log[2] + QPolyGamma[0, 1 - (I [Pi])/Log[2], 1/2])/Log[2]

In[2]:= Re[(-Log[2] + QPolyGamma[0, 1 - (I [Pi])/Log[2], 1/2])/Log[2]]

Out[2]= (-Log[2] + Re[QPolyGamma[0, 1 - (I [Pi])/Log[2], 1/2]])/Log[2]

In[3]:= N[(-Log[2] + Re[QPolyGamma[0, 1 - (I [Pi])/Log[2], 1/2]])/
Log[2]]

Out[3]= 0.7645

I also tried to consider some series of terms of functions, such as $\sum_{j=1}^{\infty}\frac{1}{1+x^j},$ hoping to estimate the sum of the desired sum, by differentiation/integration term-by-term. But perhaps it is very hard, and I was in vain. Can someone give me some clues? Many Thanks. I guess maybe we can sum the series $\sum_{j=1}^{\infty}\frac{1}{1+a^j}$ explicitly for all $a>1,$ if we can do for $a=2.$

Answer

Notice $$\frac{x}{1+x} = \frac{x(1-x)}{1-x^2} = \frac{x(1+x)-2x^2}{1-x^2} = \frac{x}{1-x} - \frac{2x^2}{1-x^2}$$
Let $q = \frac12$, we have
$$\sum_{n=1}^\infty

\frac{1}{1+2^n} =
\sum_{n=1}^\infty \frac{q^n}{1+q^n}
= \sum_{n=1}^\infty \frac{q^n}{1-q^n} - 2\sum_{n=1}^\infty \frac{q^{2n}}{1-q^{2n}}
= L(q) - 2L(q^2)
$$
where $L(x)$ is a Lambert series
defined by
$$L(x) \stackrel{def}{=} \sum_{n=1}^\infty \frac{x^n}{1-x^n}
= \frac{\psi_{x}(1) + \log(1-x)}{\log x}
$$

This Lambert series can be expressed in terms of q-polygamma function $\psi_\beta(x)$.
The sum we want becomes

$$\sum_{n=1}^\infty \frac{1}{1+2^n}
= \frac{\log(3/2) - \psi_{1/2}(1) + \psi_{1/4}(1)}{\log 2}$$

By throwing the command (Log[3/2] - QPolyGamma[1,1/2] + QPolyGamma[1,1/4])/Log[2] to WA,
one find
$$\sum_{n=1}^\infty \frac{1}{1+2^n}
\approx 0.76449978034844420919131974725549848...$$