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Some other summation formulas can be obtained by differentiating the above equations on both sides.

Assume $|r|<1$. Show that $$ a+2ar+3ar^2+\cdots = \frac{a}{(1-r)^2} $$ by starting with the geometric series formula. This seems to be trivial to prove by differentiation of both sides of the infinited geometric series formula. Is this a legal operation?

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Differentiation of both sides is a legal operation. – Matt Groff Apr 5 '13 at 14:37
For $|r|<1$ geometric series converges uniformly, hence interchange of differentiation/integration and summation are correct – Alex Apr 5 '13 at 14:43

Interchanging the order of differentiation and infinite summation is not valid in general, but it can be validated under certain circumstances.

For example, assume the power series $\sum a_n x^n$ has the radius of convergence $R > 0$. Then on the open interval $(-R, R)$, you can freely interchange the order of differentiation/integration and infinite summation: on $|x| < R$,

  • $\displaystyle \frac{d}{dx} \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n \frac{dx^n}{dx} = \sum_{n=1}^{\infty} n a_n x^{n-1}. $
  • $\displaystyle \int_{0}^{x} \sum_{n=0}^{\infty} a_n t^n \, dt = \sum_{n=0}^{\infty} a_n \int_{0}^{x} t^{n} \, dt = \sum_{n=0}^{\infty} \frac{a_n}{n+1} x^{n+1}. $

Note that the radius of convergence of the geometric series

$$ a + ax + ax^2 + ax^3 + \cdots $$

is exactly $R = 1$ unless $a = 0$. In particular, your operation is totally legal.

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If you're concerned with infinite terms, what you can do instead is differentiate finite terms, and then take the limit as $n \rightarrow \infty$. So begin with

$\displaystyle\sum_{k=0}^{n}{ar^k} = \dfrac{a(1-r^n)}{1 - r}$

Differentiating this with respect to $r$ gives us,

$\displaystyle\sum_{k=1}^{n} kar^{k-1} = \dfrac{a}{(1-r)^2}\big((1-r)(-nr^{n-1}) + (1 - r^n)\big)$

Taking the limit of both sides, and assuming that $|r|<1$ finishes the proof.

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