Find the sum: $\displaystyle \sum_{n = 1}^{\infty}\frac{1}{\left(3n-1\right)\sqrt{2}\left(4n-1\right)}$

Printable View

- Dec 23rd 2009, 07:09 PMPQREvaluate this sum.
Find the sum: $\displaystyle \sum_{n = 1}^{\infty}\frac{1}{\left(3n-1\right)\sqrt{2}\left(4n-1\right)}$

- Dec 23rd 2009, 07:45 PMVonNemo19
- Dec 24th 2009, 08:25 PMPQR
- Dec 25th 2009, 03:17 AMmr fantastic
- Dec 25th 2009, 10:25 AMKrizalid
$\displaystyle \frac{1}{(3n+2)(4n+3)}=\frac{3(4n+3)-4(3n+2)}{(3n+2)(4n+3)}=\frac{3}{3n+2}-\frac{4}{4n+3},$ 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 $\displaystyle n=0.$)