Constructing a Power Series using a geometric series as a base.

**bkbowser** · December 4th 2012, 10:28 PM

i am to use $\frac{1}{1-x}=\sum{x^n}$

to find a power series representation for $\frac{1+x}{1-x}$

additionally I need to find the interval of convergence.

since, $\frac{1}{1-x}=\sum{x^n}$

and

$(1+x)\frac{1}{1-x}=\frac{1+x}{1-x}$

then

$(1+x)*\sum{x^n}$

the book says $1+2\sum{x^n}$ is a power series representation of the original function $\frac{1+x}{1-x}$ can someone point out my error?

**emakarov** · December 5th 2012, 03:12 AM

$(1+x)*\sum{x^n}$ is correct, but it is not in the form $\sum c_nx^n$ .

**bkbowser** · December 5th 2012, 09:48 AM

we'll i don't think i can use the distributive property to make it
$\sum{x^n}+\sum{x^{n+1}}$
since this is basically an polynomial of infinitely many terms being multiplied by a binomial.

i don't really see anything to do here.

**bkbowser** · December 5th 2012, 10:10 AM

i don't know of a way to dump out the parentheses and make a " $c_n$ "

**emakarov** · December 5th 2012, 10:30 AM

Originally Posted by bkbowser

we'll i don't think i can use the distributive property to make it
$\sum{x^n}+\sum{x^{n+1}}$
since this is basically an polynomial of infinitely many terms being multiplied by a binomial.

This depends on the level of rigor with which you need to "find" the series for $\frac{1+x}{1-x}$ . Do you need an accurate proof that it converges to the function inside the interval of convergence? I agree that infinite sums require special care, but using distributivity in the way you did is fine as the first approximation.

**skeeter** · December 5th 2012, 10:32 AM

$\frac{1+x}{1-x}$

$\frac{1}{1-x} + \frac{x}{1-x}$

$(1 + x + x^2 + x^3 + ... ) + (x + x^2 + x^3 + x^4 + ...)$

$1 + 2x + 2x^2 + 2x^3 + ...$

$1 + 2(x + x^2 + x^3 + ... ) = 1 + 2\sum_{n=1}^\infty x^n$

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