How to obtain the sum of the series $\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}$?

Hint. One may prove that $$ \sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}=\sum_{n=0}^{\infty}\int_0^1\frac{x^{3n}}{2^{n}}dx=\int_0^1\sum_{n=0}^{\infty}\left(\frac{x^3}{2}\right)^{n}dx=\int_0^1\frac{2}{2-x^3}\:dx $$ Hope you can take it from here.

Observe that

$$f(x):=\sum_{n=0}^\infty\frac{x^n}{3n+1}\implies f'(x)=\sum_{n=0}^\infty\frac{nx^{n-1}}{3n+1}$$ and $$3xf'(x)+f(x)=\sum_{n=0}^\infty\frac{3n+1}{3n+1}x^n=\frac1{1-x}.$$

The solution of the homogeneous part of this linear ODE is $$f_h(x)=\frac c{\sqrt[3]x},$$

and by variation of the constant

$$f(x)=\frac1{\sqrt[3]x}\left(\int\frac{\sqrt[3]x}{1-x}dx+c\right).$$

For all $|x|<1$ we have the elementary geometric series $$\sum_{n\geq 0}x^n=\frac{1}{1-x} \\\underbrace{=}^{x\to x/2}\sum_{n\geq 0}\frac{x^n}{2^n} =\frac{2}{2-x}$$ now replacing $x$ by $x^3$ and then on integration from $0$ to $1$ we have $$\sum_{n\geq 0}\frac{1}{2^n(3n+1)}=\int_0^1\frac{2}{2-x^3}dx$$ since $x^3-2=(x-\sqrt[3]2)(x^2-\sqrt[3]{2}x+\sqrt[3]{8})$ by partial fraction of the integrand we write the last expression as $$-\int_0^1\frac{2}{x^3-2} dx=-\frac{2}{3\sqrt[3]4}\int_0^1\left(\color{red}{\frac{1}{x-\sqrt[3]{2}}}-\color{blue}{\frac{x+\sqrt[3]{16}}{x^2+\sqrt[3]{2}x +\sqrt[3]4}}\right)dx$$ It's is easy to see that red integral $$\int_0^1\color{red}{\frac{1}{x-\sqrt[3]2}}dx =\ln(|x- \sqrt[3]2|)\bigg|_0^1=\ln(|1-\sqrt[3]2|)-\ln\sqrt[3]2\cdots(1)$$ Further note that $$\int_0^1\color{blue}{\frac{x+\sqrt[3]{16}}{x^2+\sqrt[3]{2}x+\sqrt[3]{4}}}dx=\frac{1}{2}\int_0^1\left(\frac{2x+\sqrt[3]2}{x^2+\sqrt[3]2 x+\sqrt[3]4}+\frac{3\cdot \sqrt[3]{2}}{x^2+\sqrt[3]2x+\sqrt[4]{4}}\right)dx$$ The last two integral are standard and elementary logarithm and arctangent integrals and integrating them we have $$\frac{1}{2}\ln(x^2+\sqrt[3]2x+\sqrt[3]4)+\sqrt[3]{2}\tan^{-1}\left(\frac{\sqrt[3]{4}x+1}{\sqrt 3}\right)\bigg|_0^1=\frac{\ln(1+\sqrt[3]{2}+\sqrt[3]{4})-\ln(\sqrt[3]{4})}{2}+\sqrt{3}\tan^{-1}\left(\frac{\sqrt[3]{4}+1}{\sqrt{3}}\right)-\frac{\pi}{2\sqrt{3}}\cdots(2)$$ subtract $(1)$ from $(2)$ and multiply by the factor $-\frac{2}{3\sqrt[3]{2}}$ and simplification gives us the desired result of the series.

$$\frac{2}{3\sqrt[3]{4}}\left(\sqrt{3}\tan^{-1}\left(\frac{\sqrt[3]{4}+1}{\sqrt 3}\right)+\frac{1}{2}\ln(1+\sqrt[3]{2}+\sqrt[3]{4})-\ln|\sqrt[3]{2}-1| -\frac{\pi}{2\sqrt{3}}\right)\approx 1.18143\cdots$$