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?
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 ...
Merry Christmas from Italy
$\displaystyle \chi$ $\displaystyle \sigma$
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 $\displaystyle x$.
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= ___".