Radius of convergence and 2^1/2

[bedi]

Homework Statement

Suppose c_n is the digit in the nth place of the decimal expansion of 2^1/2. Prove that the radius of convergence of $\sum{c_n x^n}$ is equal to 1.

Homework Equations

The Attempt at a Solution

What I want to show is that limsup |c_n|^1/n = 1. Clearly for any c_n, (c_n)^1/n ≥ 1. So we have sup (c_n)^1/n ≥ 1 for any n. It is also easy to see that sup (c_n)^1/n ≤ n^1/n for n large enough, as c_n is at most 9. So by the Sandwich lemma limsup |c_n|^1/n = 1.

Correct?

 

[jbunniii]

What I want to show is that limsup |c_n|^1/n = 1. Clearly for any c_n, (c_n)^1/n ≥ 1.

Why is this clearly true? Are you saying that the decimal expansion of $\sqrt{2}$ doesn't contain any zero digits?

So we have sup (c_n)^1/n ≥ 1 for any n.

Yes, this part is true, because there are infinitely many nonzero $c_n$'s. (Proof?)

[Edit]: Be careful about the notation. I think you meant
$$\sup \{(c_m)^{1/m} : m \geq n\} \geq 1 \textrm{ for any } n$$
and therefore
$$\limsup (c_n)^{1/n} \geq 1$$

It is also easy to see that sup (c_n)^1/n ≤ n^1/n for n large enough, as c_n is at most 9. So by the Sandwich lemma limsup |c_n|^1/n = 1.

Correct?

Yes, that looks OK, assuming you have already proved that $\lim_{n\rightarrow \infty} n^{1/n} = 1$.