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?

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- December 13th 2009, 04:04 PMkatielawFind 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 AMchisigma
The 'infinite sum' of functions...

(1)

... does converge for all real values of the variable. For is...

(2)

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

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

(3)

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

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

Merry Christmas from Italy

- December 14th 2009, 05:10 AMkatielaw
I don't understand..Is that it? Where do I go from that?

- December 14th 2009, 05:22 AMProve It
I don't really understand your question.

Is this an infinite series?

Is it supposed to be, for example

?

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

Note that this can be rewritten as

.

Can you see that it is a geometric series, with and ?

Geometric series converge when .

So this function converges when

Solve for . - December 14th 2009, 05:28 AMHallsofIvy
A "power series", a series of the form has a "radius of convergence" but other kinds of series may not.

In particular, you can think of the series you are given, as a**geometric series**, with and .

That's chisigma's point.

Now, you have probably learned, long ago, that the geometric series will converge as long as |r|< 1. That's where chisigma gets . In fact, since is never negative, which is equivalent to or . That's true for all x except ___ and the region of convergence is "all x except x= ___". - December 14th 2009, 05:29 AMchisigma
Summarizing...

... converges ...

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

Merry Christmas

- December 14th 2009, 02:56 PMProve It
- December 14th 2009, 05:30 PMkatielaw
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 AMkatielaw
So is it the same as a sequence of functions as it would be for a power series?