Determine exactly $\displaystyle\sum_{i=0}^\infty (-1)^i \dfrac{1}{3i+1}$.
From the question it is evident that this sum has an exact value. It is also alternating and since $a_n=\dfrac{1}{3n+1}$ is a positive, decreasing sequence that approaches $0$ as $n\to \infty$ the series converges. I think the Taylor series of $\ln (1+x) = x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dots$ might be useful here, but I can't seem to make much progress from that. It seems like the sum can be expressed in terms of $\pi$.
Hint: substitute $x \to -x^3$ into the geometric series and then integrate from 0 to 1.