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

$\displaystyle a_n=a^{n^2}$ for some constant $\displaystyle a \in (0, \infty).$
So far I have this:

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

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

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

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

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 $\displaystyle x \in \mathbb{R}$) 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 $\displaystyle |a^{n^2}| \leq |a^n|$
This is only true if a>1 !

Then this inequality
Therefore: $\displaystyle |\sum a^{n^2}x^n|=|1+a^4x^2+a^9x^3+... |\leq | \sum a^{n}x^n|=|1+(ax)^2+(ax)^3+...|$
is false
I don't know for the rest

3. ah, okay then.

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

Hence:

$\displaystyle n^2 ln(a)<0$

$\displaystyle ln(a)<0$

hence $\displaystyle a<1$.

Therefore the radius of convergence is 1.

$\displaystyle ie. \ R=1$.

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