Let \(a_n \neq 0\) and assume that \(\displaystyle \lim_{n \rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} = L\). Show that \(\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{|a_n|} = L. %p_n^{\frac{1}{n}} = L.\) In particular, this shows that when applicable, the ratio test can be used to calculate the radius of convergence of a power series.
Solution
First suppose \(0<L<\infty\). Set \[ r_k=\frac{|a_{k+1}|}{|a_k|}. \] Then \(r_k\to L\), and \[ |a_n|=|a_0|\prod_{k=0}^{n-1}r_k. \] Taking logarithms, \[ \frac1n\log|a_n| =\frac{\log|a_0|}{n} +\frac1n\sum_{k=0}^{n-1}\log r_k. \] Since \(\log r_k\to\log L\), Cesàro convergence gives \[ \frac1n\log|a_n|\to\log L. \] Exponentiating yields \[ |a_n|^{1/n}\to L. \]
If \(L=0\), then for every \(\varepsilon>0\) we eventually have \(r_k<\varepsilon\), so \(|a_n|^{1/n}\le C^{1/n}\varepsilon^{1-o(1)}\) and hence the limit is \(0\). If \(L=\infty\), apply the same argument to \(1/|a_n|\), whose successive ratio tends to \(0\). Thus in all cases \[ \boxed{\lim_{n\to\infty}|a_n|^{1/n}=L.} \]