Find the sum: $\displaystyle \sum_{n = 1}^{\infty}\frac{1}{\left(3n-1\right)\sqrt{2}\left(4n-1\right)}$
$\displaystyle \frac{1}{(3n+2)(4n+3)}=\frac{3(4n+3)-4(3n+2)}{(3n+2)(4n+3)}=\frac{3}{3n+2}-\frac{4}{4n+3},$ now use $\displaystyle \frac1{n+1}=\int_0^1x^n\,dx$ and the expansion for the geometric series. Solve the remaining integral.
(By the way, the general term changes because the series starts from $\displaystyle n=0.$)