Differentiating a sum to infinity

Question

Let $G:[0,1] → R$ be given by:

$G(s)= \sum s^n$ (this sum is from $0$ to $\infty$)

I have to evaluate $G'(s)$ in two different ways:

Differentiate the terms in the sum terms by term.
Sum the series on the right hand side of equation and then differentiate the sum.

Then by equating the two answers show that:

$\sum ns^{n-1}$ (this sum is from $1$ to $\infty$) = $\frac {1}{(1-s)^2}$

I feel silly for not knowing exactly how to do this, anyway what I have done is:

$\sum ns^{n-1}$ (this sum is from $1$ to $\infty$)
Differential of $\frac {1}{1-s}$ is $\frac {1}{(1-s)^2}$

The way to show ranges on your sums is to show $\sum_{n=0}^\infty$ You get that in $\LaTeX$ with \sum_{n=0}^\infty — Ross Millikan, Oct 30 '13 at 15:52

score 2 · Answer 1 · answered Oct 30 '13 at 15:50

2

$\frac{1}{1-s}=(1-s)^{-1}$. Differentiate with the chain rule, as $(-1)(1-s)^{-2}(-1)=\frac{1}{(1-s)^2}$. One -1 comes from the exponent, one comes from the derivative of $1-s$.

answered Oct 30 '13 at 15:50

vadim123

82,796

Of course, silly error. Thanks. – Rich A Oct 30 '13 at 15:53

Differentiating a sum to infinity

1 Answers1