How to prove $\sum_{k=1}^∞ k r^{k-1} = \frac{(1)}{(1-r)^2}$

Question

My textbook asks me to prove that

when 1>|r|,

$$\sum_{k=1}^∞ k r^{k-1} = \frac{(1)}{(1-r)^2}$$

Does anyone know if this can be proven?

Note: This question has been flagged as duplicate, but before you go on look at other answers, here is how I solved this question:

$\sum_{k=1}^n k r^2 = 1+2r+3r^2+...... $

$S= 1+2r+3r^2+4r^3+.....$

$Sr= r+2r^2+3r^3+4r^4+....$

$S-Sr= 1+r+r^2+r^3+r^4+....$

$S-Sr= \frac{1}{1-r}$

$S(1-r)\frac{1}{(1-r)}=\frac{(1)}{(1-r)(1-r)}$

$S=\frac{(1)}{(1-r)^2}$

Yes, I know. My computation reached me at: k/(1-r) That is why I am asking. Am I doing something wrong? — Onur, Feb 24 '17 at 09:53

score 2 · Answer 1 · answered Feb 24 '17 at 09:52

2

Compute for $|r|<1$

$ \frac{1}{(1-r)^2}=(\sum_{k=0}^{\infty}r^k)^2 $

with Cauchy product or differentiate $f(r):=\sum_{k=0}^{\infty}r^k \quad (=\frac{1}{1-r})$

answered Feb 24 '17 at 09:52

Fred

score 1 · Answer 2 · answered Feb 24 '17 at 09:54

Hint:

We know that for $|x|<1$ $$\frac {1}{1-x} = 1+ x + x^2 + x^3 +\cdots \tag {1}$$

Now differentiating both sides with respect to $x$, $$\frac {\mathrm {d}}{\mathrm {d}x}(1) \implies ? $$

Hope you can take it from here.

2 Answers2