Radius of Convergence of power series anx^n^2

[looserlama]

Homework Statement

Suppose that the power series $\sum$a_nxⁿ for n=0 to n=∞ has a radius of convergence R$\in$(0,∞). Find the radii of convergence of the series $\sum$a_nxⁿ² from n=0 to n=∞ and $\sum$a_nx²ⁿ.

Homework Equations

Radius of convergence theorem:

R = 1/limsup|a_n|^1/n is the radius of convergence of the series $\sum$a_n(x-c)ⁿ

Root Test:

Let L = limsup|a_n|^1/n, then

If L<1, the series $\sum$a_n converges absolutely
If L>1, the series diverges
If L=1, then the test is inconclusive.

The Attempt at a Solution

So we know that limsup|a_n|^1/n = $\frac{1}{R}$.

So for the second series ($\sum$a_nx²ⁿ) I think this is easy:

limsup|a_n|^1/n|x|^2n/n = limsup|a_n|^1/n|x|² = |x|²/R < 1 So that the series converges

Therefore, |x|² < R $\Rightarrow$ |x| < $\sqrt{R}$

So R₂ = $\sqrt{R}$ (Where R₂ is the r.o.c. of the second series)So I tried doing this with the first series, but it doesn't work.

So then I showed that limsup|b_n|^1/n2 < 1 implies that $\sum$b_n from n=0 to n=∞ converges absolutely (basically a modification of the root test)

So then if I use this for the first series:

limsup|a_n|^1/n2|x|^n2/n2 = limsup|a_n|^1/n2|x| < 1 So that it converges.

Therefore |x| < 1/limsup|a_n|^1/n2 = R₁

But I'm assuming they want it in terms of R, so this doesn't really work...

I feel like I'm close I'm just missing something?

Any help would be greatly appreciated.

 

[Dick]

I'm a little bit dubious about some this, and I think you meant your initial series to be a_n*x^n, not a_n*x^(n^2), but let's assume all of your limits exist. In the x^(2n) case you got |x|^2<R. In the x^(n^2) case you will get limit n->infinity |x|^n<R. What does that tell you about radius of convergence?

 

[looserlama]

How can we have |x|^n <R if n->infinity?

Wouldn't it be 1/R*limsup|x|^n <1 ?

So limsup|x|^n < R and wouldn't that only be true for |x| < 1 ?

 

[Dick]

How can we have |x|^n <R if n->infinity?

Wouldn't it be 1/R*limsup|x|^n <1 ?

So limsup|x|^n < R and wouldn't that only be true for |x| < 1 ?

Yes, I think you need |x|<1.

 

[looserlama]

Oh ok, it just doesn't seem like a very formal way of showing it. Is there not a better way of doing it?

 

[Dick]

Pick a number b such that |b|<1 and apply the root test to $a_{n} b^{n^2}$. Now do the same for |b|>1.