2

Find the sum of $\sum_{n=1}^{\infty} n5^{2n-2}x^{2n+1}$

I am wondering how one would be able to solve this series?

janmarqz
  • 10,538
DragonDude
  • 71
  • 1
    Begin by differentiating the geometric series and its formula. You can rearrange this series into one that matches the form necessary. – Clayton Nov 12 '15 at 19:33
  • By changing parameters I get $\frac{1}{125} \sqrt{z}\sum_{n=1}^{\infty}nz^n$. Can it help? – Hoseyn Heydari Nov 12 '15 at 19:50

1 Answers1

1

Hint. You may start from the useful finite evaluation: $$ 1+r+r^2+...+r^n=\frac{1-r^{n+1}}{1-r}, \quad |r|<1. \tag1 $$ Then by differentiating $(1)$ you get $$ 1+2r+3r^2+...+nr^{n-1}=\sum_{k=1}^{n-1}kr^k=\frac{1-r^{n+1}}{(1-r)^2}+\frac{-(n+1)r^{n}}{1-r}, \quad |r|<1. \tag2 $$

Then, using $|r|<1$, let $n \to \infty$ in $(2)$ and put $r:=5^2x^2$ in the obtained identity.

Olivier Oloa
  • 120,989