Evaluate this sum.

**PQR** · December 23rd 2009, 07:09 PM

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

**VonNemo19** · December 23rd 2009, 07:45 PM

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.

**PQR** · December 24th 2009, 08:25 PM

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.

**mr fantastic** · December 25th 2009, 03:17 AM

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.

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.

**Krizalid** · December 25th 2009, 10:25 AM

$\frac{1}{(3n+2)(4n+3)}=\frac{3(4n+3)-4(3n+2)}{(3n+2)(4n+3)}=\frac{3}{3n+2}-\frac{4}{4n+3},$ 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 $n=0.$ )

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