Sum of $\sum_{k=0}^{\infty} k \rho^{k-1}$

Question

I encounter the infinite series $\sum_{k=0}^{\infty} k \rho^{k-1}$ in a math textbook where the answer is directly given to be $\dfrac{1}{(1-\rho)^2}$ when $|\rho| < 1$. However, I don't understand how to obtain this result. Could someone give me a hint of how to approach this problem?

score 0 · Answer 1 · answered Dec 25 '18 at 20:02

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Start off with the geometric sum$$\frac 1{1-x}=\sum\limits_{n\geq0}c^n$$Now differentiate it with respect to $x$ to get$$\frac 1{(1-x)^2}=\sum\limits_{n\geq0}n x^{n-1}$$

answered Dec 25 '18 at 20:02

Frank W

5,897

score 0 · Accepted Answer · answered Dec 25 '18 at 20:02

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Hint: Think of the convergent series $$\sum_{k=0}^\infty \rho^k=\frac{1}{1-\rho}, \qquad |\rho|<1,$$ and differentiate both sides with respect to $\rho$ (why is interchanging differentiation and summation possible?).

answered Dec 25 '18 at 20:02

I looked at this post and understand now why we can interchange differentiation and summation. Thanks a lot! – ark Dec 25 '18 at 20:29

score 0 · Answer 3 · answered Dec 25 '18 at 20:03

If you are familiar with the formula and derivation of the geometric series, then you can justify this by showing you can pass derivatives into a series inside the radius of convergence. This post contains a good rundown of why what we are about to do is valid.

Recall the following: $$ \sum_{k=0}^\infty x^k = \frac{1}{1-x} \quad \text{when } |x|<1 $$ If you differentiate both sides you find the following: $$ \sum_{k=0}^\infty kx^{k-1} = \frac{1}{(1-x)^2} \quad \text{when } |x|<1 $$

score 0 · Answer 4 · answered Dec 25 '18 at 20:07

Here is a method which avoids use of derivatives or integrals. Let $S$ be the sum of the series. By distributing $\rho$ among the summands of your series, show that $\rho S = S - \sum_{k = 0}^\infty \rho^k$. Use the formula for the sum of a geometric series and solve for $S$ to obtain $S = \dfrac{1}{(1 - \rho)^2}$.

score 0 · Answer 5 · answered Dec 25 '18 at 20:07

This is an arithmetic-geometric series. One term $(k)$ is in arithmetic progression, while the other $(\rho^{k-1})$ is in geometric progression. It converges for $|\rho|<1$. You can find the value like this:

$\displaystyle S=\sum_{k=1}^{\infty}k\rho^{k-1}=1+2\rho+3\rho^2+4\rho^3...$

$\displaystyle\rho S=\sum_{k=1}^\infty k\rho^k=\rho+2\rho^2+3\rho^3+4\rho^4...$

$\displaystyle (1-\rho)S=1+\rho+\rho^2...=\frac1{1-\rho}$ for $|\rho|<1$

$\displaystyle\therefore S=\frac1{(1-\rho)^2}$

Sum of $\sum_{k=0}^{\infty} k \rho^{k-1}$

5 Answers5