How to calculate $\sum_{j=0}^\infty (-1)^j/(3j + 1)$ How do you calculate the value of 
$$\sum_{j=0}^\infty \frac{(-1)^j}{3j + 1} =  1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+ \cdots $$
I know this series will converge .. but I don't know how to compute the limit.
 A: Hint. One may observe that
$$
\sum_{j=0}^\infty \frac{(-1)^j}{3j + 1}=\sum_{j=0}^\infty\int_0^1(-1)^jx^{3j}dx=\int_0^1\frac{dx}{1+x^3}=\frac{\pi}{3\sqrt{3}}+\frac{1}{3}\cdot\ln 2
$$ where we have used the partial fraction decomposition
$$
\frac{1}{1+x^3}=\frac{-x+2}{3 \left(x^2-x+1\right)}+\frac{1}{3(x+1)}.
$$
A: Hint:
We need $$\sum_{r=0}^\infty\dfrac{(-1)^r}{3r+1}$$
Now $$\sum_{r=0}^\infty\dfrac{(x^3)^r}{3r+1}=\dfrac1x\int\sum_{r=0}^\infty x^{3r}dx$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{j = 0}^{\infty}{\pars{-1}^{\,j} \over 3j + 1} & =
\sum_{j = 0}^{\infty}\pars{{1 \over 6j + 1} - {1 \over 6j + 4}} =
{1 \over 6}\sum_{j = 0}^{\infty}\pars{{1 \over j + 1/6} - {1 \over j + 2/3}}
\\[5mm] & =
{1 \over 6}\pars{H_{-1/3} - H_{-5/6}}\qquad\pars{~H_{z}:\ Harmonic\ Number~}
\end{align}

The Harmonic Numbers are evaluated with
  Gauss Digamma Theorem:

$$
\left\{\begin{array}{rcl}
\ds{H_{-1/3}} & \ds{=} &
\ds{-\gamma + {\root{3} \over 6}\,\pi - {3 \over 2}\,\ln\pars{3}}
\\[2mm]
\ds{H_{-5/6}} & \ds{=} &
\ds{-\gamma - {\root{3} \over 2}\,\pi - {3 \over 2}\,\ln\pars{3} - 2\ln\pars{2}}
\\[3mm]
&&\gamma:\ Euler\!-\!Mascheroni\ Constant
\end{array}\right.
$$

$$
\bbx{\sum_{j = 0}^{\infty}{\pars{-1}^{\,j} \over 3j + 1} =
{\root{3} \over 9}\,\pi + {1 \over 3}\,\ln\pars{2}}
$$
