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

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!

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  • $\begingroup$ This may give you some ideas. $\endgroup$ Commented Apr 23, 2014 at 14:16

2 Answers 2

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

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  • $\begingroup$ 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? $\endgroup$
    – kanbhold
    Commented Apr 23, 2014 at 14:53
  • $\begingroup$ 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? $\endgroup$ Commented Apr 25, 2014 at 14:51
  • $\begingroup$ 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? $\endgroup$ Commented Apr 25, 2014 at 14:53
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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.

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