Series problem with limiting sum

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June 8th 2007, 01:44 AM
jacs

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Series problem with limiting sum

A limit problem i have tried for hours now and just cannot wrap my head around
Solution is apparently (1 + x)/(1 - x)^2

any help greatly appreciated,
thanks
jacs
June 8th 2007, 06:17 AM
ThePerfectHacker

We are adding all of these together. So let us add them in a convenient way.

First we add the left-most coloums together:
$1+x+x^2+x^3+... = \frac{1}{1-x}$

Second we add the next left-most coloums together:
$2x+2x^2+2x^3+... = 2x(1+x+x^2+...) = \frac{2x}{1-x}$

Third we add the next left-most colums together:
$2x^2+2x^3+2x^4+... = 2x^2(1+x+x^2+...) = \frac{2x^2}{1-x}$

And so on....

So in total we need to add,
$\frac{1}{1-x} + \frac{2x}{1-x}+\frac{2x^2}{1-x}+\frac{2x^3}{1-x}+...$

I am going to add and subtract $\frac{1}{1-x}$ to make it an easier form.

$\left(\frac{2}{1-x} + \frac{2x}{1-x}+\frac{2x^2}{1-x}+\frac{2x^3}{1-x}+... \right) - \frac{1}{1-x}$
Factor again,
$\frac{2}{1-x}\left( 1+x+x^2+x^3+... \right) - \frac{1}{1-x}$

$\frac{2}{(1-x)^2} - \frac{1}{1-x} =\boxed{ \frac{1+x}{(1-x)^2}}$
June 8th 2007, 06:22 AM
Soroban

Hello, jacs!

Quote:

Observe that:
. . $1 \:=\:1$
. $3x\:=\:x + 2x$
$5x^2\:=\:x^2 + 2x^2 + 2x^2$
$7x^3 \:=\:x^3 + 2x^3 + 2x^3 + 2x^3$
$9x^4 \:=\:x^4 + 2x^4 + 2x^4 + 2x^4 + 2x^4$

By studying the arrangement, or otherwise, find in simplest algebraic form,
. . an expression for the limit of the series: . $1 + 3x + 5x^2 + 7x^3 + 9x^4 + \cdots$

Solution: . $\frac{1+x}{(1-x)^2}$

Add the stack of equations . . .

The left side is: . $S\:=\:1 + 3x + 5x^2 + 7x^3 + 9x^4 + \cdots$

The right side is: . $(1 + x + x^2 + \cdots) + (2x + 2x^2 + 2x^3 + \cdots) + (2x^2 + 2x^3 + 2x^4+\cdots) + \cdots$

. . . . . . . . . . . $= \;(1 + x + x^2 + \cdots) + 2x(1 + x + x^2 + \cdots) + 2x^2(1 + x + x^2 + \cdots) + \cdots$

Factor: . $(1 + x + x^2 + \cdots)\cdot[1 + 2x + 2x^2 + 2x^3 + \cdots]$

. . . . $= \;(1 + x + x^2 + \cdots)\cdot\left[1 + 2x(1 + x + x^2 + \cdots)\right]$

The expressions in parentheses are geometric series:
. . first term $a = 1$ , common ratio $r = x$
Their sum is: . $\frac{1}{1-x}$

Hence, we have: . $S \;=\;\frac{1}{1-x}\cdot\left[1 + 2x\left(\frac{1}{1-x}\right)\right]$

. . which simplifies to: . $\boxed{S \;=\;\frac{1+x}{(1-x)^2}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Another solution . . . (They did say "otherwise", right?)

We are given: . $S \;=\;1 + 3x + 5x^2 + 7x^3 + 9x^4 + \cdots$

Multiply by $x\!:\;xS \;=\;\qquad x + 3x^2 + 5x^3 + 7x^4 + \cdots$

Subtract: . $S - xS \;=\;1 + 2x + 2x^2 + 2x^3 + 2x^4 + \cdots$

. . . . . . . $(1 - x)S \;=\; 1 + 2x\underbrace{(1 + x + x^2 + x^3 +\cdots)}_{\text{geometric series}}$

So we have : . $(1 - x)S \;=\;1 + 2x\left(\frac{1}{1-x}\right) \;=\;\frac{1+x}{1-x}$

Therefore: . $S \;=\;\frac{1+x}{(1-x)^2}$
June 8th 2007, 06:39 AM
jacs

thanks to both of you. i dont think i ever woudl have figured that one out in a milion years. Going ot have a study of all the different methods and see which one i can try to replicate without giving myself and annureism.

thanks, you guys rock!