How to derive the closed form of the sum of $kr^k$

Question

$$ \sum_{k=0}^{n}kr^k = r\frac{1-(n+1)r^n + nr^{n+1}}{ (1 - r)^2 } $$

How to derive it? I read about some finite calculus, and i understand how to tackle sums of $x^2$, $x^3$, etc.. But I don't know if the same methods can be used on this sum?

Answer 1

You can prove the relation using induction:

Basis (n=0): clear.

Inductive step: $$ r\frac{1 - (n+1)r^{n} + nr^{n+1}}{(1-r)^2} + (n+1)r^{n+1} = r\frac{1-(n+1)r^{n} + nr^{n+1} + (n+1)r^{n} - 2(n+1)r^{n+1} + (n+1)r^{n+2} }{(1-r)^2} = r\frac{1-(n+2)r^{n+1} + (n+1)r^{n+2}}{(1-r)^2} $$

Then, it follows. Another way to come up with it is differentiating the regular formula for the geometric progression:

$$ \frac{d}{dr}\sum_{k=0}^{n} r^{k} = \sum_{k=0}^{n} kr^{k-1} = \frac{d}{dr}\frac{1-r^{n+1}}{1-r} = \frac{-(n+1)r^{n}(1-r) + 1 - r^{n+1}}{(1-r)^2} = \frac{1-(n+1)r^{n} + nr^{n+1}}{(1-r)^2} $$

Then, just multiply the whole thing by $r$ and you're done.

Answer 2

Hint: $kr^k = r k r^{k-1} = r \frac{d}{dr}r^{k}$, now interchange summation and derivative.

Answer 3

This is an arithmetico-geometric series.

$$\begin{align} \sum_{k=0}^{n}kr^k=S&=0+1\cdot r+2r^2+3r^3+4r^4+\cdots+nr^n \\rS&=\qquad \qquad \quad r^2+2r^3+3r^4+\cdots+(n-1)r^n+nr^{n+1}\\ \text{Subtracting:}\\ (1-r)S&=\qquad \quad r+\ \ r^2+\ \ r^3+\ r^4 +\cdots+\ \ r^n \qquad \quad -nr^{n+1}\\ &=\frac{r(1-r^n)}{1-r}-nr^{n+1}\\ S&=\frac {r(1-r^n)-(1-r)nr^{n+1}}{(1-r)^2}\\ &=\frac{r(1-r^n)-r(1-r)nr^n}{(1-r)^2}\\ &=r\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}\qquad \blacksquare\end{align}$$

Answer 4

Via summation by parts:

$$\sum_{k=0}^n f_k g_k = f_n G_n - \sum_{j=0}^{n-1} \left( f_{j+1}- f_j\right) G_j,\ \text{where}\ G_j = \sum_{k=0}^j g_k$$

Here we have: $f_k=k$, $g_k=r^k$, $G_j=\sum\limits_{k=0}^j r^k=\dfrac{r^{j+1}-1}{r-1}$, $f_{j+1}-f_j=\left(j+1\right)-j=1$.

Then $\sum\limits_{k=0}^{n} kr^k=nG_n-\sum\limits_{j=0}^{n-1} 1\cdot G_j$. But $$\sum\limits_{j=0}^{n-1} G_j=\sum\limits_{j=0}^{n-1}\dfrac{r^{j+1}-1}{r-1}=\dfrac{1}{r-1}\left(\sum\limits_{j=0}^{n-1}r^{j+1}-n\right)=\dfrac{1}{r-1}\left(\dfrac{r^{n+1}-r}{r-1}-n\right).$$

Finally we have $\sum\limits_{k=0}^{n} kr^k=n\cdot\dfrac{r^{n+1}-1}{r-1}-\dfrac{1}{r-1}\left(\dfrac{r^{n+1}-r}{r-1}-n\right)=\dfrac{nr^{n+2}-(n+1)r^{n+1}+r}{\left(r-1\right)^2}\blacksquare$.