# Find the region of convergence

• December 13th 2009, 04:04 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?
• December 14th 2009, 12:53 AM
chisigma
The 'infinite sum' of functions...

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

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

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

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

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

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

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

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

Merry Christmas from Italy

$\chi$ $\sigma$
• December 14th 2009, 05:10 AM
katielaw
I don't understand..Is that it? Where do I go from that?
• December 14th 2009, 05:22 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

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

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

Note that this can be rewritten as

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

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

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

So this function converges when

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

Solve for $x$.
• December 14th 2009, 05:28 AM
HallsofIvy
A "power series", a series of the form $\sum a x^n$ has a "radius of convergence" but other kinds of series may not.

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

Now, you have probably learned, long ago, that the geometric series $\sum ar^n$ will converge as long as |r|< 1. That's where chisigma gets $|\frac{1}{1+x^2}|< 1$. In fact, since $1+ x^2$ is never negative, $\frac{1}{1+x^2}< 1$ which is equivalent to $1+ x^2> 1$ or $x^2> 0$. That's true for all x except ___ and the region of convergence is "all x except x= ___".
• December 14th 2009, 05:29 AM
chisigma
Summarizing...

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

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

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

Merry Christmas

$\chi$ $\sigma$
• December 14th 2009, 02:56 PM
Prove It
Quote:

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

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

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

Actually $\frac{1}{1 + x^2} \leq 1$.
• December 14th 2009, 05:30 PM
katielaw
It is a series of functions. I thought the f_n(x) indicated that. I'm sorry I wasn't clearer.
• December 15th 2009, 10:34 AM
katielaw
So is it the same as a sequence of functions as it would be for a power series?