1. ## series problem

I am having difficulty solving this math problem:
Suppose it is known that |x| < 1. What then is the sum from n=1 to n=infinity of nx^n?

I know how to do it if there is a constant in front of the x, but I don't know what to do since there's a variable in front of it.
Anybody have any ideas, it's really starting to annoy me?

Thanks a Lot
-Nertil

2. Originally Posted by nertil1
I am having difficulty solving this math problem:
Suppose it is known that |x| < 1. What then is the sum from n=1 to n=infinity of nx^n?

I know how to do it if there is a constant in front of the x, but I don't know what to do since there's a variable in front of it.
Anybody have any ideas, it's really starting to annoy me?

Thanks a Lot
-Nertil
*Warning*-This is not a formal proof but it captures the idea.

Let,
$S=x+2x^2+3x^3+4x^4+...$
Thus,
$xS=x^2+2x^3+3x^4+4x^5+...$
Let,
$S'=x+x^2+x^3+x^4+...$
Thus,
$xS+S'=x+2x^2+3x^3+4x^4+...$
Thus,
$xS+S'=S$
Thus,
$\frac{S'}{1-x}=S$
But,
$S'=\frac{x}{1-x}$ because it is an infinite geometric series,
Thus,
$S=\frac{x}{(1-x)^2}$

3. I was thinking about another way of doing this problem.
Notice the partial sums,
$S_1=x$
$S_2=x+2x^2=(x+x^2)+x^2=x\left(\frac{1-x^2}{1-x}\right)+x^2$
$S_3=x+2x^2+3x^3=(x+x^2+x^3)+(x^2+x^3)+x^3$= $x\left(\frac{1-x^3}{1-x}\right)+x^2\left(\frac{1-x^2}{1-x}\right)+x^3$
In general,
$S_n=\frac{1}{1-x}[x(1-x^n)+x^2(1-x^{n-1})+...+x^n]$
All of this is a manipulation of the geometric series,

As you take the limit the exponents die,
Thus,
$S_{\infty}=\frac{1}{1-x}[x+x^2+x^3+..]$
But that is the geometric series thus,
$S_{\infty}=\frac{x}{(1-x)^2}$

This proof is more formal.

4. I realized yet one more way, but this is super informal.
Integrate part by part,
$x+2x^2+3x^3+4x^4+...$
Thus,
$\frac{x^2}{2}+\frac{2x^3}{3}+\frac{3x^4}{4}+...$
Express as,
$(-x+x)+\left(-\frac{x}{2}+x^2\right)+\left(-\frac{x^2}{3}+x^3\right)+...$
Express as,
$\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-...\right)+(x+x^2+x^3+...)$
Know the expansion for infinite series we have,
$\ln|1-x|+\frac{x}{1-x}$
Thus, the derivative of this must be the initial series
$-\frac{1}{1-x}+\frac{1}{(1-x)^2}=\frac{x}{(1-x)^2}$
----------------

5. how did you get an S prime in the first solution?

6. Originally Posted by nertil1
how did you get an S prime in the first solution?
As $|x|<1$ consider:

$
\frac{1}{1-x}=1+x+x^2+...+x^n+...
$

Now differentiate both sides:

$
\frac{1}{(1-x)^2}=1+2x+3x^2+...+nx^{n-1}+...
$
,

so multiplying through by $x$ gives:

$
\frac{x}{(1-x)^2}=x+2x^2+3x^3+...+nx^n+...
$

You might want to worry about convergence, if you check you should find
the rules for term by term differentiation are satisfied.

RonL