Let an>0 for n∈N. The convergence radius of the series ∑∞n=0anzn is 1q with q=lim or q = \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}, if these limits exist. Therefore the \lims must be identical.
However I was wondering whether there exists a more elementary way to show this identity? (Or generally any other way?)
Answer
The general relation is
\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} \le \liminf_{n \to \infty} \sqrt[n]{\mathstrut a_n} \le \limsup_{n \to \infty} \sqrt[n]{\mathstrut a_n} \le \limsup_{n \to \infty} \frac{a_{n+1}}{a_n} \quad \text{(*)}
provided that all a_n are positive. It follows that if
\lim_{n \to \infty} \frac{a_{n+1}}{a_n} exists then
\lim_{n \to \infty} \sqrt[n]{\mathstrut a_n} exists as well and they are equal.
To prove the rightmost inequality, define S := \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}. For every \epsilon > 0 there is a N \in \mathbb N
such that
\frac{a_{n+1}}{a_n} < S + \epsilon \quad \text{ for } n \ge N \, .
\implies a_n < a_N \, (S + \epsilon)^{n - N} \quad \text{ for } n \ge N
\implies \sqrt[n]{\mathstrut a_n} < (S + \epsilon) \sqrt[n]{ a_N (S + \epsilon)^{-N}}
\implies \limsup_{n \to \infty} \sqrt[n]{\mathstrut a_n} \le S + \epsilon \, .
The leftmost inequality can be proved in the same way, or by taking
the reciprocals.
That the convergence radius R of a power series \sum_{n=0}^\infty a_n z^n can be
determined with the root test or
with the ratio test
is a consequence of the above relation (*), not the other way around.
But note that the tests are slighty different:
\frac 1R = \limsup_{n \to \infty} \sqrt[n]{\mathstrut |a_n|}
holds even if the limit does not exists.
\frac 1R = \lim_{n \to \infty} \bigl| \frac{a_{n+1}}{a_n} \bigr|
is only valid if the limit on the right-hand side exists. You cannot
generally replace the \lim by \limsup in the ratio test.
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