# Problem 30

• Jul 9th 2007, 12:34 PM
ThePerfectHacker
Problem 30
Show that $\displaystyle \sum_{n=1}^{\infty}nx^{n-1} = \frac{1}{(1-x)^2}$ for $\displaystyle |x|<1$ without differenciating the geometric power series.
• Jul 9th 2007, 12:55 PM
red_dog
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
ThePerfectHacker
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
Rebesques
Quote:

I am not sure if is a fair question to ask but I post it anyway.
That's ok, we just apply the relevant theorem :p:p
• Jul 11th 2007, 09:46 AM
ThePerfectHacker
Quote:

Originally Posted by Rebesques
That's ok, we just apply the relevant theorem :p:p

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.