Radius of convergence

**Showcase_22** · February 21st 2009, 04:08 AM

Find the radius of convergence of the power series $\sum a_nx^n$ for the following choices of $(a_n; n \geq 0).$

$a_n=a^{n^2}$ for some constant $a \in (0, \infty).$

So far I have this:

$\sum a^{n^2}x^n=1+a^4x^2+a^9x^3+......$

but $|a^{n^2}| \leq |a^n|$

Therefore: $|\sum a^{n^2}x^n|=|1+a^4x^2+a^9x^3+... |\leq | \sum a^{n}x^n|=|1+(ax)^2+(ax)^3+...|$

This will only converge if $a<1$ . Hence the radius of convergence of $| \sum a^{n}x^n|=|1+(ax)^2+(ax)^3+...|$ is 1 (ie. $R=1$ ).

So I have an upper bound, but I can't find a lower bound. Since x can be negative (i'm assuming from the question that $x \in \mathbb{R}$ ) I can't just make the original power series greater than 1!

Any help would be great!

**Moo** · February 21st 2009, 04:13 AM

Hello,

Originally Posted by Showcase_22

but $|a^{n^2}| \leq |a^n|$

This is only true if a>1 !

Then this inequality

Therefore: $|\sum a^{n^2}x^n|=|1+a^4x^2+a^9x^3+... |\leq | \sum a^{n}x^n|=|1+(ax)^2+(ax)^3+...|$

is false
I don't know for the rest

**Showcase_22** · February 21st 2009, 05:35 AM

ah, okay then.

Well how about this:

$a^{n^2}<1$ for the power series to converge.

Hence:

$n^2 ln(a)<0$

$ln(a)<0$

hence $a<1$ .

Therefore the radius of convergence is 1.

$ie. \ R=1$ .

I would have to prove that the series diverges for $a \geq 1$ but I think I can do that.

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