Integration to series

Guess a particular solution I(n) which is in the form of an integral with lower and upper limits 0 and 1 respectively, to the difference equation,

I(n+1) + I(n) = 1/(3n+1) for all n>=0

Hence, calculate the infinite series:
∑[k=0 to infinity] (-1)^k /(3k+1)

[Hint: Integrate[from 0 to 1] x^(3n) dx = 1/(3n+1)]