anonymous 5 years ago Find the value of Sigma n=1 to infinity n/k^n for k>1

1. watchmath

You already asked this since yesterday. Could you please write down the expression using the nice math symbol so everybody can read the problem correctly?

2. anonymous

$\[\sum_{n=1}^{\infty}n/k ^{n}$

3. anonymous

for k>1... I believe k can be replaced with x

4. watchmath

$\frac{x}{(1-x)^2}=\sum_{n=1}^\infty nx^{n}$ Then to compute your series, just substitute $$x=\frac{1}{k}$$ to the left hand side :D.

5. anonymous

So would it look like -2/k$-4/k^2-6/k^3-8/k^4$?