What is the sum of $\sum\limits_{i=1}^{n}ip^i$ and does it matter, for finite n, if $|p|>1$ or $|p|<1$ ?

Edition :

Why can I integrate take sum and then take the derivative ? I think that kind of trick is not always allowed.

ps. I've tried this approach but I made mistake when taking derivative, so I've asked, mayby I should use some program (or on-line app) for symbolic computation.

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    $\begingroup$ What did you try? $\endgroup$ – Did Aug 8 '12 at 6:46
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    $\begingroup$ Hint: Try to integrate your sum. $\endgroup$ – Seirios Aug 8 '12 at 6:47
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    $\begingroup$ Multiply by $p$ and subtract. $\endgroup$ – PAD Aug 8 '12 at 6:48
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    $\begingroup$ These questions are not exactly the same as your question, but some answers gives a more general answers there, so you can find answer to your question there (or you can try to use a similar method and a make the generalization yourself). 1 2 3 $\endgroup$ – Martin Sleziak Aug 8 '12 at 6:50
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    $\begingroup$ If you think your methods are "not allowed" ... use them anyway to get a formula, then prove the formula by induction. $\endgroup$ – GEdgar Aug 8 '12 at 14:16

$$\sum_{k=1}^n kp^k=\frac{p(np^{n+1}-(n+1)p^n+1)}{(p-1)^2}\tag{*}$$

Proof by induction:

  1. $n=1$: $ p=\frac{p(p^{1+1}-(1+1)p^1+1)}{(p-1)^2}=\frac{p(p^{2}-2p+1)}{(p-1)^2}=p$

  1. $$ \begin{eqnarray} (n+1)p^{n+1}+\sum_{k=1}^n kp^k&=&(n+1)p^{n+1} + \frac{p(np^{n+1}-(n+1)p^n+1)}{(p-1)^2}\\ &=&\frac{(n+1)p^{n+1}(p^2-2p+1)}{(p-1)^2} + \frac{p(np^{n+1}-(n+1)p^n+1)}{(p-1)^2}\\ &=&\frac{np^{n+3}+p^{n+3}-np^{n+2}-2p^{n+2}+p}{(p-1)^2}\\ &=&\frac{p((n+1)p^{n+2}-(n+1+1)p^{n+1}+1)}{(p-1)^2}\\ &=&\sum_{k=1}^{n+1} kp^k \end{eqnarray} $$

If $p=1$, we expect $\sum_{k=1}^n k\cdot 1^k= \frac12 n(n+1)$: Since the RHS of $(*)$ gives $\frac00$, when we insert $p=1$, we apply L'Hospital's rule two times: $$ \lim_{p\to 1} \frac{p(np^{n+1}-(n+1)p^n+1)}{(p-1)^2} =\lim_{p\to 1} \frac{n(n+2)p^{n+1}-(n+1)^2p^{n}+1}{2(p-1)}\\ =\frac12 \lim_{p\to 1} (n(n+1)(n+2)p^{n}-n(n+1)^2p^{n-1})\\ =\frac12 n(n+1) \underbrace{\lim_{p\to 1} p^{n-1}((n+2)p-(n+1))}_{=1}\\ $$ If $n\to \infty$, the series converges due to ratio test ($\lim_{n\to \infty}\left|\frac{(k+1)}{k}p\right|<1$), when $|p|<1$. You'll get $$ \sum\limits_{k=1}^\infty kp^k = \frac p{(p-1)^2}+\underbrace{\lim_{n\to \infty} \frac{np^{n+2}-(n+1)p^{n+1}}{(p-1)^2}}_{=0} = \frac p {(p-1)^2} $$


Start with $f(p)=\sum_{i=1}^n p^i=\frac {p(p^n-1)}{p-1}$

your answer is $\ pf'(p)$ (compute it both ways...)


Here is yet another way: $$ \begin{align*} \sum_{i=1}^n ip^i &= \sum_{i=1}^n \sum_{j=i}^n p^j \\ &= \sum_{i=1}^n \frac{p^{n+1}-p^i}{p-1} \\ &= n\frac{p^{n+1}}{p-1} - \frac{1}{p-1} \sum_{i=1}^n p^i \\ &= n\frac{p^{n+1}}{p-1} - \frac{p^{n+1}-1}{(p-1)^2}. \end{align*} $$


I like calculus. My first line would be

$ Q(p) + \sum\limits_{i=1}^{n}p^i = \sum\limits_{i=1}^{n}(i+1)p^i = \sum\limits_{i=1}^{n} \frac{d}{dp} p^{i+1} = \frac{d}{dp} \sum\limits_{i=1}^{n} p^{i+1} $

(Writing $Q(p)$ for your function $\sum\limits_{i=1}^{n}ip^i$)

That was the old swap the sum and the differentiation trick. It remains to the sum the geometric series and differentiate it.

Since we used calculus, we can use the same method will also work for the continuous problem $ R(p) = \int\limits_{i=1}^{n}ip^i $


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