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Question:

Suppose a power series $$\sum_{n=0}^\infty a_n x^n$$ Satisfies $$a_{n-2} + (n^2 + \alpha^2)a_n =0,\ for\ all\ n\geqslant 2$$ What is the radius of convergence of the power series?

I have tried:

1) Split into 2, odds and even

2) $$a_{0} + (n^2 + \alpha^2)a_2=0$$ $$a_{1} + (n^2 + \alpha^2)a_3=0$$ 3)$$a_{0} + (n^2 + \alpha^2)a_2=a_{1} + (n^2 + \alpha^2)a_3 $$ 4) $$\frac {a_0-a_1}{a_3-a_0} = (n^2 + \alpha^2)$$ 5) I am stuck here , I don't know whether this approach is correct.

Bernard
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winson
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1 Answers1

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After thinking a bit about it, I think there is simpler.

We have $|a_{n-2}| =(n^2+\alpha^2)|a_n|\ge n^2|a_n|$ thus $|a_n|\le \frac 1{n^2}|a_{n-2}|\le \dfrac{|a_{n\bmod 2}|}{(n!!)^2}$

If we take $M=\max(|a_0|,|a_1|)$ we have $|a_n|\le \dfrac{M}{(n!!)^2}$ and this gives absolute convergence with an infinite radius

In fact $\alpha$ is just a disturbance here, $a_n\to 0$ fast enough with the division by $n^2$ each step.

zwim
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