# Thread: Sum of a series

1. ## Sum of a series & maclin series

So I have written down some notes, and I have forgotten why i wrote what i wrote down.

Find the sum of the series
$\sum_{n=1}^\infty nx^n$

I wrote down these steps

$\sum_{n=1}^\infty nx^n$= x $\sum_{n=1}^\infty nx^{n-1}$=x* $\frac{1}{(1-x)^2}$= $\frac{x}{(1-x)^2}$ for abs(x)<1

I don't know how they come about...

2. Use $\sum nx^{n-1} = \frac{\mathrm{d}}{\mathrm{d}x} \sum x^n$.

Use $\sum nx^{n-1} = \frac{\mathrm{d}}{\mathrm{d}x} \sum x^n$.
Edit: Never mind...

4. Originally Posted by Prove It
Edit: Never mind...
:[

Use $\sum nx^{n-1} = \frac{\mathrm{d}}{\mathrm{d}x} \sum x^n$.
Maddas is right. Than take the derivative of $\frac{1}{1-x}$. Since $\frac{1}{1-x}=\sum x^n$ for |x|<1.

:[
OK, I'll fill in the steps then :P

$\sum_{n = 1}^{\infty}nx^n = x\sum_{n =1}^{\infty}nx^{n - 1}$.

Notice that $\frac{d}{dx}\left(\sum_{n = 1}^{\infty}x^n\right) = \sum_{n = 1}^{\infty}nx^{n - 1}$.

Also notice that $\sum_{n = 1}^{\infty}x^n$ is a geometric series with $a = x$ and $r = x$.

So $\sum_{n = 1}^{\infty}x^n = \frac{x}{1 - x}$ for $|x| < 1$.

Therefore $\frac{d}{dx}\left(\sum_{n =1}^{\infty}x^n\right) = \frac{d}{dx}\left(\frac{x}{1 - x}\right)$

$= \frac{1}{(1- x)^2}$.

So this means

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

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

$\sum_{n = 1}^{\infty}nx^n = \frac{x}{(1-x)^2}$, provided of course that $|x| < 1$.

7. :D

8. they're using power series as functions..got it

thanks for the help there!

I have another question though if you don't mind...Don't want to make another topic

For the maclin series of $\sqrt{1+x}$, after the limit ratio test is

$\frac{abs(x)}{2}$ $\lim_{n\to\infty}\frac{2n-1}{n+1}$ = abs(x)<1. Hence 1 is the radius of convergence. I understand how most of everything comes about, but what i don't understand is why 2n-1 is there. since the original was 1*3*5...(2n-3), shouldn't that be (2n-1)*(2n-2)?