Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

2 Answers

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.