Let \(a_n\neq 0\) and show that \[ \lim_{n\to \infty} {\abs{a_{n+1}} \over \abs{a_n}} = L \implies \lim_{n\to\infty} \abs{a_n}^{1\over n} = L .\] In particular, this shows that when applicable, the ratio test can be used to calculate the radius of convergence of a power series.
Solution
1. Put \(r_n = \abs{a_{n+1}}/\abs{a_n}\). Then \(r_n>0\) and \(r_n \to L\).
The quotient is defined and positive because every \(a_n\neq0\); its limit is \(L\) by hypothesis.
2. For any \(\varepsilon > 0\), put \(\ell=\max\{L-\varepsilon,0\}\). There is \(N\) such that \(\ell \le r_n < L + \varepsilon\) for all \(n \ge N\).
Since \(r_n\to L\), there is \(N\) with \(L-\varepsilon<r_n<L+\varepsilon\) for \(n\ge N\). Together with \(r_n>0\) from step 1, this gives \(\ell\le r_n\).
3. For \(n > N\),\[\abs{a_N}\,\ell^{n-N} \le \abs{a_n} < \abs{a_N}(L + \varepsilon)^{n-N}.\]
The product telescopes: \(\abs{a_n} = \abs{a_N} \prod_{k=N}^{n-1} r_k\). Bound each factor \(r_k\) by step 2.
4. One has\[L - \varepsilon \le \ell \le \liminf_{n\to\infty} \abs{a_n}^{1/n} \le \limsup_{n\to\infty} \abs{a_n}^{1/n} \le L + \varepsilon.\]
Take \(n\)th roots in step 3 and let \(n\to\infty\). Since \(a_N\neq0\), \(\abs{a_N}^{1/n} \to 1\), and \(c^{(n-N)/n} \to c\) for every \(c>0\). If \(\ell=0\), the lower bound is \(0\le\liminf\abs{a_n}^{1/n}\).
5. One has \(\lim_{n\to\infty} \abs{a_n}^{1/n} = \boxed{L}\).
Step 4 holds for every \(\varepsilon > 0\). Letting \(\varepsilon\to0\) gives \(L\le\liminf\abs{a_n}^{1/n}\le\limsup\abs{a_n}^{1/n}\le L\).
6. Q.E.D.
Step 5 is the required limit.