Sum of this eries: $\sum_{k=1}^{\infty}kp(1-p)^{k-1}$

Question

$\sum_{k=1}^{\infty}kp(1-p)^{k-1}$

Can someone help me evaluate this sum? I couldn't even start, I have just written down the first couple of elements, but didn't help either.

Thanks!

This may give you some ideas. – David Mitra Apr 23 '14 at 14:16 — David Mitra, Apr 23 '14 at 14:16

score 2 · Answer 1 · answered Apr 23 '14 at 14:23

2

Can you compute the sum $$\sum_{k=1}^\infty (1-p)^k?$$ Now what happens if you differentiate term-by-term?

answered Apr 23 '14 at 14:23

Edvard Fagerholm

I can, that is a geometric series. And the term-by-term derivatives of this are the terms of my original problem. If I just differentiate the sum of this do I get the sum of my problem? – kanbhold Apr 23 '14 at 14:53
You get something very close. $-\sum_{k=1}^\infty k(1-p)^{k-1}$. What do you have to do to get the original and how is that related to what the geometric series in my answer converges to? – Edvard Fagerholm Apr 25 '14 at 14:51
Lookup the theorem about differentiating term-by-term. Hint: Yes, you can differentiate the expression $1/p$ and get $-\sum_{k=1}^\infty k(1-p)^{k-1}=-1/p^2$... Based on what I just wrote what must the answer then be? – Edvard Fagerholm Apr 25 '14 at 14:53

score 1 · Answer 2 · answered Apr 23 '14 at 14:29

1

let the sum begin at k=0 and write instead of $kp(1-p)^k$ the expression $(k+1)p(1-P)^k$. Then expand the expression behind the sigma sign.

greetings,

calculus.

answered Apr 23 '14 at 14:29

callculus42

2 Answers2