Show that $\displaystyle \sum_{n=1}^{\infty}nx^{n-1} = \frac{1}{(1-x)^2}$ for $\displaystyle |x|<1$withoutdifferenciating the geometric power series.

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- Jul 9th 2007, 12:34 PM #1

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- Jul 9th 2007, 12:55 PM #2
Let $\displaystyle S_n=\sum_{k=1}^nkx^{k-1}$

We have $\displaystyle S_n=1+2x+3x^2+\ldots +nx^{n-1}$ (1).

Multiplying (1) with $\displaystyle x$ we have

$\displaystyle xS_n=x+2x^2+3x^3+\ldots +(n-1)x^{n-1}+nx^n$ (2)

Substracting (2) from (1) yields

$\displaystyle S_n(1-x)=1+x+x^2+\ldots +x^{n-1}-nx^n=\frac{1-x^n}{1-x}-nx^n\Rightarrow S_n=\frac{1-x^n}{(1-x)^2}-\frac{nx^n}{1-x}$

Then $\displaystyle \sum_{n=1}^{\infty}nx^{n-1}=\lim_{n\to\infty}S_n=\frac{1}{(1-x)^2}$

- Jul 10th 2007, 03:12 PM #3

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Since you answered so quickly, here is a second challenge. I am not sure if is a fair question to ask but I post it anyway.

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Having shown that $\displaystyle \sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2}$

Can you prove the following?

Let $\displaystyle f(x) = \sum_{n=0}^{\infty} a_nx^n$ with radius of convergence $\displaystyle R>0$

Then, $\displaystyle f$ is differenciable on $\displaystyle (-R,R)$ and furthermore $\displaystyle f'(x) = \sum_{n=1}^{\infty} na_nx^{n-1}$.

Note: This works even if $\displaystyle x$ is a complex number, but the proofs are completely anagolous.

- Jul 11th 2007, 09:36 AM #4

- Jul 11th 2007, 09:46 AM #5

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I am not sure what you mean. In my Analysis class we proved differenciation term-by-term by working backwards with the Fundamental Theorem of Calculus after proving integration term-by-term. And that was just Real fucntions.

By my Complex Analysis book had a really nice approach to this proof. Without using more advandeced techiqnues. It was able to prove term-by-term differenciation by using the series posted above. It was a long but straightforward derivation. That is what I am asking to show.