# Evaluate this sum.

• Dec 23rd 2009, 08:09 PM
PQR
Evaluate this sum.
Find the sum: $\sum_{n = 1}^{\infty}\frac{1}{\left(3n-1\right)\sqrt{2}\left(4n-1\right)}$
• Dec 23rd 2009, 08:45 PM
VonNemo19
Quote:

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

Try partial fraction decomposition. See if you can obtain a telescoping sum.
• Dec 24th 2009, 09:25 PM
PQR
Quote:

Originally Posted by VonNemo19
Try partial fraction decomposition. See if you can obtain a telescoping sum.

I have: $\frac{3}{2\sqrt{2}}-\frac{4}{3\sqrt{2}}+\frac{3}{5\sqrt{2}}-\frac{4}{7\sqrt{2}}+\frac{3}{8\sqrt{2}}-\frac{4}{11\sqrt{2}}+\frac{3}{11\sqrt{2}}-\frac{4}{15\sqrt{2}}+...$

Not sure what to do with it, though. (Thinking)
• Dec 25th 2009, 04:17 AM
mr fantastic
Quote:

Originally Posted by PQR
I have: $\frac{3}{2\sqrt{2}}-\frac{4}{3\sqrt{2}}+\frac{3}{5\sqrt{2}}-\frac{4}{7\sqrt{2}}+\frac{3}{8\sqrt{2}}-\frac{4}{11\sqrt{2}}+\frac{3}{11\sqrt{2}}-\frac{4}{15\sqrt{2}}+...$

Not sure what to do with it, though. (Thinking)

Actually the problem is tougher than suggested by the first glance. I don't have time now - I will try to reply with some suggestions later if no-one else has done so.
• Dec 25th 2009, 11:25 AM
Krizalid
$\frac{1}{(3n+2)(4n+3)}=\frac{3(4n+3)-4(3n+2)}{(3n+2)(4n+3)}=\frac{3}{3n+2}-\frac{4}{4n+3},$ now use $\frac1{n+1}=\int_0^1x^n\,dx$ and the expansion for the geometric series. Solve the remaining integral.

(By the way, the general term changes because the series starts from $n=0.$)