Deriving the ratio test from the nth root test for convergence I am following the derivation on the first page of this:
https://www.maa.org/sites/default/files/0025570x33450.di021200.02p0190s.pdf
I am getting caught up on one of the details: In asserting that $\lim_{n\rightarrow \infty}a_0=1$, I think $a_0$ is being treated as a constant, in which case I agree. But then, I think, taking the limit of the product:
  $\lim_{n\rightarrow \infty}|\frac{a_{n}}{a_{n-1}}*\frac{a_{n-1}}{a_{n-2}}*....*
  \frac{a_{1}}{a_{0}}|^{1/n} \lim_{n\rightarrow \infty}|a_{0}|^{1/n}=
  \lim_{n\rightarrow \infty}|\frac{a_{n}}{a_{n-1}}*\frac{a_{n-1}}{a_{n-2}}*....*
  \frac{a_{1}}{a_{0}}|^{1/n}$ is suggestive of the conclusion that if the limit of the ratio tends to some constant $L$, then the limit of the nth root of this product will also converge to $L$. Am I right in this conclusion or missing a step?
edit: I am hoping to conclude that if $\lim_{n \rightarrow \infty}|\frac{a_{n}}{a_{n+1}}|=R$, then using the nth root test which states that if $\lim_{n\rightarrow \infty}\sup|a_{n}|^{1/n}=1/R$ then the series $\sum_{n=0}^{\infty} a_{n}z^{n}$ converges with radius R, we know that the same series converges with radius R. But I am using that the product of limits is the limit of products if they exist, which seems to be fudging indexing a little: I am using that the ratios are dependent on n when I conclude this, but not when i assert that the $\lim_{n\rightarrow \infty}|a_{0}|^{1/n}=1$

In the paragraph following that section, they're saying the root test is stronger than the ratio test, because if the ratios approach a certain limit, then the average (geometric mean) of the ratios will approach the same limit.
The geometric mean is simply multiplying the numbers and taking the n-th root, so it sounds like your interpretation is correct.
