For which values of x does this series converge?

[moxy]

Homework Statement

For what real values of x does this series converge?

\sum_{n=1}^{∞}{\frac{1}{(\frac{1}{x})^n + x^n}}

The Attempt at a Solution

I've rewritten the series as \sum_{n=1}^{∞}{\frac{x^n}{1 + x^{2n}}} and I know that I can make each term larger, say
{\frac{x^n}{1 + x^{2n}}} < {\frac{x^n}{x^{2n}}} = \left({\frac{x}{x^{2}}}\right)^n = \left(\frac{1}{x}\right)^n

Then this will converge, by comparison to the geometric series, when |x| > 1. But surely this can't be a method for finding ALL values of x for which the original series converges, can it? Because I changed the series , so all I've found is the values of x for which \sum_{n=1}^{∞}{\frac{x^n}{x^{2n}}} converges. How can I go about this in a different way?

Then there's the question of uniform convergence. Can I just use the M-test for both parts, and say that

\left|\frac{x^n}{1 + x^{2n}}\right| < \left|\frac{x^n}{x^{2n}}\right| = \left|\frac{1}{x}\right|^n = M_n

so, \sum_{n=1}^{∞}{M_n} = \sum_{n=1}^{∞} \left|\frac{1}{x}\right|^n converges uniformly when |x| > 1?

I just fear that I'm simplifying things too much and not getting ALL of the possible values of x such that the series converges. Any suggestions?

 

[lanedance]

this problem might be well suited to a comparison

say 0<x<1, then for all n
<br /> \frac{1}{x^n} + x^n &gt; \frac{1}{x^n}<br />

Hence
<br /> \frac{1}{\frac{1}{x^n} + x^n}&lt; \frac{1}{\frac{1}{x^n}}<br />

note you could do similar thing for x>1 and negative values

 

[moxy]

Okay, that makes sense. But since I'm only comparing to other series, can I be sure that I'm catching all of the values of x where the original series converges?

EDIT: Nevermind...since I'm comparing the original series based on specific intervals of x, I'll be fine.

So I'll get the same interval for convergence and uniform convergence by the M-test?

 

[lanedance]

not quite

consider x=1/2, using the fact below do you think the series will converge?
<br /> \frac{1}{\frac{1}{(\frac{1}{2})^n} + (\frac{1}{2})^n}&lt; \frac{1}{\frac{1}{(\frac{1}{2})^n}}= \frac{1}{2^n}<br />


now consider x=2, using the fact below do you think the series will converge?
<br /> \frac{1}{\frac{1}{(2)^n} + (2)^n}&lt; \frac{1}{2^n}<br />

note the symmetry in arguments, for any x>1, you can use the same argument above for y=1/x, with 0<y<1