Convergence of a series of functions

**tbyou87** · November 3rd 2007, 12:19 PM

Hi,

Show that $\sum \frac{x^n}{1+x^n}$ converges for x in [0,1).

I'm not really sure how to get this one started.

Thanks

**Plato** · November 3rd 2007, 12:28 PM

Is this true?
$x \in [0,1)\quad \Rightarrow \quad \frac{x}{{1 + x}} \le x$

**tbyou87** · November 3rd 2007, 12:44 PM

Originally Posted by Plato

Is this true?
$x \in [0,1)\quad \Rightarrow \quad \frac{x}{{1 + x}} \le x$

Yes, it is true. I think that implies

$\frac{x^n}{1+x^n} \le x^n$

Then I guess you just prove $\sum x^n$ converges for x in [0,1).

So B = lim sup |an|^(1/n) = lim sup |1|^1/n = 1. So radius of convergence R = 1/1 = 1. Thus converges for (-1,1). Since [0,1) is a subset of (-1,1), I should converge in that interval.

Is this valid?

**Jhevon** · November 3rd 2007, 09:17 PM

Originally Posted by tbyou87

Yes, it is true. I think that implies

$\frac{x^n}{1+x^n} \le x^n$

Then I guess you just prove $\sum x^n$ converges for x in [0,1).

So B = lim sup |an|^(1/n) = lim sup |1|^1/n = 1. So radius of convergence R = 1/1 = 1. Thus converges for (-1,1). Since [0,1) is a subset of (-1,1), I should converge in that interval.

Is this valid?

you are working too hard, note that $\sum x^n$ is a geometric series. what are the restraints on x for convergence?

**tbyou87** · November 3rd 2007, 09:23 PM

o yeah x element (-1,1).
Thanks forgot about that lol.

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