# series problem

• Mar 6th 2006, 02:50 PM
nertil1
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
• Mar 6th 2006, 06:12 PM
ThePerfectHacker
Quote:

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,
$\displaystyle S=x+2x^2+3x^3+4x^4+...$
Thus,
$\displaystyle xS=x^2+2x^3+3x^4+4x^5+...$
Let,
$\displaystyle S'=x+x^2+x^3+x^4+...$
Thus,
$\displaystyle xS+S'=x+2x^2+3x^3+4x^4+...$
Thus,
$\displaystyle xS+S'=S$
Thus,
$\displaystyle \frac{S'}{1-x}=S$
But,
$\displaystyle S'=\frac{x}{1-x}$ because it is an infinite geometric series,
Thus,
$\displaystyle S=\frac{x}{(1-x)^2}$
• Mar 6th 2006, 06:38 PM
ThePerfectHacker
I was thinking about another way of doing this problem.
Notice the partial sums,
$\displaystyle S_1=x$
$\displaystyle S_2=x+2x^2=(x+x^2)+x^2=x\left(\frac{1-x^2}{1-x}\right)+x^2$
$\displaystyle S_3=x+2x^2+3x^3=(x+x^2+x^3)+(x^2+x^3)+x^3$=$\displaystyle x\left(\frac{1-x^3}{1-x}\right)+x^2\left(\frac{1-x^2}{1-x}\right)+x^3$
In general,
$\displaystyle S_n=\frac{1}{1-x}[x(1-x^n)+x^2(1-x^{n-1})+...+x^n]$
All of this is a manipulation of the geometric series,

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

This proof is more formal.
• Mar 6th 2006, 06:58 PM
ThePerfectHacker
I realized yet one more way, but this is super informal.
Integrate part by part,
$\displaystyle x+2x^2+3x^3+4x^4+...$
Thus,
$\displaystyle \frac{x^2}{2}+\frac{2x^3}{3}+\frac{3x^4}{4}+...$
Express as,
$\displaystyle (-x+x)+\left(-\frac{x}{2}+x^2\right)+\left(-\frac{x^2}{3}+x^3\right)+...$
Express as,
$\displaystyle \left(-x-\frac{x^2}{2}-\frac{x^3}{3}-...\right)+(x+x^2+x^3+...)$
Know the expansion for infinite series we have,
$\displaystyle \ln|1-x|+\frac{x}{1-x}$
Thus, the derivative of this must be the initial series :eek:
$\displaystyle -\frac{1}{1-x}+\frac{1}{(1-x)^2}=\frac{x}{(1-x)^2}$
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• Mar 7th 2006, 07:59 AM
nertil1
how did you get an S prime in the first solution?
• Mar 7th 2006, 08:16 AM
CaptainBlack
Quote:

Originally Posted by nertil1
how did you get an S prime in the first solution?

As $\displaystyle |x|<1$ consider:

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

Now differentiate both sides:

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

so multiplying through by $\displaystyle x$ gives:

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

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

RonL