Radius and interval of convergence for

[philnow]

Homework Statement

Hey all. I'm being asked to find the radius and interval of convergence for the series from 1 to infinity:

x^n/(1*3*5...(2n-1))

I have a feeling this is pretty straight forward (just apply the ratio test etc...), but my trouble lies in defining the denominator. Something to do with factorials? How do I get something to represent 1*3*5 all the way to *(2n-1)?

 

[Dick]

Ok, so a_n=1/(1*3*5...(2n-1)). I think that represents it pretty well. What's a_(n+1)/a_n?

 

[philnow]

a_(n+1)/a_n would be =

1*3*5...(2n-1) / 1*3*5...(2n-1)(2n+1)

correct? In this case it simplifies to 1/(2n+1)?

 

[Dick]

a_(n+1)/a_n would be =

1*3*5...(2n-1) / 1*3*5...(2n-1)(2n+1)

correct? In this case it simplifies to 1/(2n+1)?

Absolutely correct. So what does the ratio test tell you?

 

[philnow]

Convergent for all x^n I believe, so radius of convergence = inf?

 

[Dick]

Convergent for all x^n I believe, so radius of convergence = inf?

Sure. lim a_(n+1)/a_n=0. So infinite radius of convergence. Converges for all x.

 

[philnow]

Thanks very much!

 

[fmam3]

Just to nitpick on the absolutely perfect argument above --- it's technically an implication of the ratio test and not because of the ratio test. If the power series is of the form $$\sum a_n x^n$$, then we define $$\beta = \limsup |a_n|^{1/n}$$ and the radius of convergence as $$R = 1 / \beta$$.

So, technically, one is supposed to compute $$\beta$$, but often it is hard to compute the n-th root of things. But the trick is that we see this inequality relationship
$$\liminf |a_{n+1} / a_n| \leq \liminf |a_n|^{1/n} \leq \limsup |a_n|^{1/n} \leq \limsup|a_{n+1} / a_n|$$. And since if $$\lim |a_{n+1} / a_n|$$ exists, then $$\lim |a_{n+1} / a_n| = \liminf |a_{n+1} / a_n| = \limsup|a_{n+1} / a_n|$$, implying immediately that $$\limsup |a_n|^{1/n} = \lim |a_{n+1} / a_n|$$.

I'm just nitpicking :P