Is the Cauchy-Hadamard Theorem Misrepresented on Mathworld?

[quasar987]

My book says that the Cauchy-Hadamar theorem reads: Consider a power serie $\sum_{n=0}^{\infty} a_n x^n$. The radius of convergence is given by

$$r = \frac{1}{\lim_{n \rightarrow \infty} |a_n|^{1/n}}$$

But on mathworld, it is enounced as follow (notice that it's the UPPER limit).

Which is considerably different, because for the serie

$$\sum_{n=0}^{\infty}\frac{x^{2n}}{5^n}$$

for exemple, which can be rewritten as a power serie of the form

$$\sum_{n=0}^{\infty}a_n x^n$$

where a_n = 0 if n is even and a_n = 5^(-n/2) if n is odd, we have that $\lim_{n \rightarrow \infty} |a_n|^{1/n}$ does not exist because the subsequence a_n where n is even converges towards 0 and the subsequence a_n where n is odd converges towards 5^(-1/2). So the limit doesn't exist.

On the other hand, the upper limit of the sequence |a_n|^{1/n} is obviously 5^(-1/2) since |a_n| is decreasing.

So the two theorems are not equivalent.

Who's right?

 

[shmoe]

Your verson of the theorem (without the upper limit) would tell you nothing in this case since the limit does not exist how you've included the odd coefficients. However if you rearrange the series slightly:

$$\sum_{n=0}^{\infty}\frac{(x^2)^{n}}{5^n}$$

and write $$a^n=5^{-n}$$

then the limit in your version exists and says the series converges when

$$|x^2|<1/5$$

and you get the same answer for the radius of convergence.


Both are correct, the matworld version is a little more general in what it applies to since you aren't requiring the limit itself to exist, just the upper limit.

 

[quasar987]

Ooooh.

Thanks for clarifying that shmoe.