Question

Find the value of Sigma n=1 to infinity n/k^n for k>1

Answer 1

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?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

So would it look like -2/k\[-4/k^2-6/k^3-8/k^4\]?