# Find the region of convergence

• Dec 13th 2009, 03:04 PM
katielaw
Find the region of convergence
Given f_n(x) = (x^2)/(1+x^2)^n, find the region of convergence. I'm not sure what the "region" is. Is it just the same as the radius of convergence?
• Dec 13th 2009, 11:53 PM
chisigma
The 'infinite sum' of functions...

$\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{x^{2}}{(1+x^{2})^{n}}$ (1)

... does converge for all real values of the $\displaystyle x$ variable. For $\displaystyle x\ne 0$ is...

$\displaystyle |\frac{1}{1+x^{2}}|<1$ (2)

... and the (1) is a convergent 'geometric series'. For $\displaystyle x=0$ every term of (1) vanishes so that the 'infinite sum' vanishes also...

For the complex 'infinite sum' of functions...

$\displaystyle f(z) = \sum_{n=0}^{\infty} \frac{z^{2}}{(1+z^{2})^{n}}$ (3)

... the problem is a little more 'complex'... of course (Wink) ...

http://digilander.libero.it/luposabatini/notte1.jpg

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$
• Dec 14th 2009, 04:10 AM
katielaw
I don't understand..Is that it? Where do I go from that?
• Dec 14th 2009, 04:22 AM
Prove It
Quote:

Originally Posted by katielaw
Given f_n(x) = (x^2)/(1+x^2)^n, find the region of convergence. I'm not sure what the "region" is. Is it just the same as the radius of convergence?

I don't really understand your question.

Is this an infinite series?

Is it supposed to be, for example

$\displaystyle \sum_{n = 0}^\infty \frac{x^2}{(1 + x^2)^n}$?

If it is, you're trying to find the values for which the function converges.

Note that this can be rewritten as

$\displaystyle x^2 \sum_{n = 0}^\infty \frac{1}{(1 + x^2)^n}$.

Can you see that it is a geometric series, with $\displaystyle a = 1$ and $\displaystyle r = \frac{1}{1 + x^2}$?

Geometric series converge when $\displaystyle |r| < 1$.

So this function converges when

$\displaystyle \left|\frac{1}{1 + x^2}\right| < 1$

Solve for $\displaystyle x$.
• Dec 14th 2009, 04:28 AM
HallsofIvy
A "power series", a series of the form $\displaystyle \sum a x^n$ has a "radius of convergence" but other kinds of series may not.

In particular, you can think of the series you are given, $\displaystyle \sum \frac{x^2}{(1+x^2)^n}$ as a geometric series, $\displaystyle \sum ar^n$ with $\displaystyle a= x^2$ and $\displaystyle r= \frac{1}{1+ x^2}$.
That's chisigma's point.

Now, you have probably learned, long ago, that the geometric series $\displaystyle \sum ar^n$ will converge as long as |r|< 1. That's where chisigma gets $\displaystyle |\frac{1}{1+x^2}|< 1$. In fact, since $\displaystyle 1+ x^2$ is never negative, $\displaystyle \frac{1}{1+x^2}< 1$ which is equivalent to $\displaystyle 1+ x^2> 1$ or $\displaystyle x^2> 0$. That's true for all x except ___ and the region of convergence is "all x except x= ___".
• Dec 14th 2009, 04:29 AM
chisigma
Summarizing...

$\displaystyle \sum_{n=0}^{\infty}\frac{x^{2}}{(1+x^{2})^{n}}$

... converges $\displaystyle \forall x \in \mathbb {R}$ ...

http://digilander.libero.it/luposabatini/notte1.jpg

Merry Christmas

$\displaystyle \chi$ $\displaystyle \sigma$
• Dec 14th 2009, 01:56 PM
Prove It
Quote:

Originally Posted by HallsofIvy
A "power series", a series of the form $\displaystyle \sum a x^n$ has a "radius of convergence" but other kinds of series may not.

In particular, you can think of the series you are given, $\displaystyle \sum \frac{x^2}{(1+x^2)^n}$ as a geometric series, $\displaystyle \sum ar^n$ with $\displaystyle a= x^2$ and $\displaystyle r= \frac{1}{1+ x^2}$.
That's chisigma's point.

Now, you have probably learned, long ago, that the geometric series $\displaystyle \sum ar^n$ will converge as long as |r|< 1. That's where chisigma gets $\displaystyle |\frac{1}{1+x^2}|< 1$. In fact, since $\displaystyle 1+ x^2$ is never negative, $\displaystyle \frac{1}{1+x^2}< 1$ which is equivalent to $\displaystyle 1+ x^2> 1$ or $\displaystyle x^2> 0$. That's true for all x except ___ and the region of convergence is "all x except x= ___".

Actually $\displaystyle \frac{1}{1 + x^2} \leq 1$.
• Dec 14th 2009, 04:30 PM
katielaw
It is a series of functions. I thought the f_n(x) indicated that. I'm sorry I wasn't clearer.
• Dec 15th 2009, 09:34 AM
katielaw
So is it the same as a sequence of functions as it would be for a power series?