What is the sum of $\sum\limits_{i=1}^{n}ip^i$?

Question

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.

Answer 1

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...)

Answer 2

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

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$

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

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 $$ \sum\limits_{k=1}^\infty kp^k = \frac p{(p-1)^2}+\underbrace{\lim_{n\to \infty} \frac{np^{n+2}-(n+1)p^{n+1}}{(p-1)^2}}_{=0} = \frac p {(p-1)^2} $$

Answer 3

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

Answer 4

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.

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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 $