Finding the value of the infinite sum $1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... $ Can anyone help me to find what is the value of $1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... $ when it tends to infinity
The first i wanna find the pattern but it seems do not have any unique pattern can anyone help me?

Hint. Observe that the general term of your series is 
$$\frac{(-1)^n}{3n+1}. $$
Then you may write
$$
\begin{align}
\sum_{n=0}^{\infty}\frac{(-1)^n}{3n+1}&=\sum_{n=0}^{\infty}(-1)^n\int_0^1 x^{3n} dx\\\\
&=\int_0^1 \sum_{n=0}^{\infty}(-1)^nx^{3n} dx\\\\
&=\int_0^1 \frac{1}{1+x^3} dx\\\\
&=\int_0^1 \left(\frac{1}{3 (1+x)}+\frac{2-x}{3 \left(1-x+x^2\right)}\right) dx\\\\
&=\frac{1}{3}\ln 2+\frac{\pi\sqrt{3}}{9}
\end{align}
$$

Some details.
$$
\begin{align}
\int_0^1 \frac{1}{1+x^3} dx&=\frac{1}{3}\int_0^1\!\frac{dx}{(1+x)}+\frac{1}{3}\int_0^1\!\frac{2-x}{(x-1/2)^2+3/4} dx\\\\
&=\frac{1}{3}\ln 2+\frac{1}{3}\int_{-1/2}^{1/2}\!\frac{3/2-u}{u^2+3/4} du \quad (u=x-1/2, \,dx=du)\\\\
&=\frac{1}{3}\ln 2+\frac{1}{3}\int_{-1/2}^{1/2}\!\frac{3/2}{u^2+3/4} du\\\\
&=\frac{1}{3}\ln 2+\int_{0}^{1/2}\!\frac{1}{u^2+3/4} du\\\\
&=\frac{1}{3}\ln 2+\left.\frac{2}{\sqrt{3}}\arctan \left( \frac{2}{\sqrt{3}}u\right)\right|_{0}^{1/2} \\\\
&=\frac{1}{3}\ln 2+\frac{\pi\sqrt{3}}{9}
\end{align}
$$ where we have used $\displaystyle \arctan \!\left(\! \frac{1}{\sqrt{3}}\!\right)\!=\frac{\pi}{6}$.
