# Math Help - Radius of convergence

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!

2. 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

3. ah, okay then.

$a^{n^2}<1$ 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.