What is the sum of $\sum\limits_{i=1}^{n}ip^i$? 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.
 A: Here is yet another way:
$$ \begin{align*}
\sum_{i=1}^n ip^i &= \sum_{i=1}^n \sum_{j=i}^n p^j \\ &=
\sum_{i=1}^n \frac{p^{n+1}-p^i}{p-1} \\ &=
n\frac{p^{n+1}}{p-1} - \frac{1}{p-1} \sum_{i=1}^n p^i \\ &=
n\frac{p^{n+1}}{p-1} - \frac{p^{n+1}-1}{(p-1)^2}.
\end{align*}
$$
A: 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...)
A: $$\sum_{k=1}^n kp^k=\frac{p(np^{n+1}-(n+1)p^n+1)}{(p-1)^2}\tag{*}$$
Proof by induction:


*

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





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


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