# Sum of Sequence of Reciprocals of 3 n + 1 Alternating in Sign

 $\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac 1 {3 n + 1}$ $=$ $\displaystyle 1 - \frac 1 4 + \frac 1 7 - \frac 1 {10} + \frac 1 {13} - \cdots$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \dfrac {\pi \sqrt 3} 9 + \dfrac {\ln 2} 3$ $\quad$ $\quad$