<s>[INST] Closed form of $\sum_{n=0}^{\infty }\frac{1}{2^n(4n-1)}$ Closed form of $$\sum_{n=0}^{\infty }\frac{1}{2^n(4n-1)}$$
the Wolfram gave  me the answer below, but I ask if there is a clear value of this sum
 
 [/INST] Consider
$$f(x) = \sum_{n=0}^{\infty} \frac{x^{4 n}}{4 n+3} $$
$$\frac{d}{dx} \left [x^3 f(x) \right ] = \sum_{n=0}^{\infty} x^{4 n+2}  = \frac{x^2}{1-x^4}$$
Integrate both sides...
$$x^3 f(x) = \frac14 \log{\left (\frac{1+x}{1-x} \right )} - \frac12 \arctan{x} + C $$
where $C=0$ by considering $x=0$.   The series is equal to
$$f(2^{-1/4}) = \sum_{n=0}^{\infty} \frac1{2^n} \frac1{4 n+3} = 2^{3/4} \left [\frac14 \log{\left (\frac{1+2^{-1/4}}{1-2^{-1/4}} \right )} - \frac12 \arctan{2^{-1/4}} \right ]$$
The relation to your series is 
$$-1 + \frac12 f(2^{-1/4}) \approx -0.779247$$
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