- Show that if \(\{ a _ { n } \} _ { n = 0 } ^ { \infty }\) is a sequence of non-zero complex numbers such that
\[ \lim _ {n \to \infty} \frac {| a _ {n + 1} |}{| a _ {n} |} = L, \]
then
\[ \lim _ {n \to \infty} | a _ {n} | ^ {1 / n} = L. \]
In particular, this exercise shows that when applicable, the ratio test can be used to calculate the radius of convergence of a power series.
Solution
Fix \(\varepsilon>0\); if \(L>0\), take also \(\varepsilon<L\). When \(L=0\), the lower bounds below are replaced by the trivial bound \(0\le\abs{a_n}^{1/n}\), and the argument gives \(\limsup\abs{a_n}^{1/n}\le\varepsilon\).
1. There is \(N\) such that for all \(n > N\),\[|a_N| (L - \varepsilon)^{n-N} < |a_n| < |a_N| (L + \varepsilon)^{n-N}.\]
By the hypothesis there is \(N\) with \(L - \varepsilon < |a_{k+1}|/|a_k| < L + \varepsilon\) for all \(k \ge N\). For \(n>N\) the telescoping product \(|a_n| = |a_N| \prod_{k=N}^{n-1} |a_{k+1}|/|a_k|\) then gives the bounds.
2. \(L - \varepsilon \le \liminf |a_n|^{1/n} \le \limsup |a_n|^{1/n} \le L + \varepsilon\).
Taking \(n\)th roots in step 1, which preserves order of positive quantities, \[|a_N|^{1/n} (L - \varepsilon)^{(n-N)/n} < |a_n|^{1/n} < |a_N|^{1/n} (L + \varepsilon)^{(n-N)/n}.\] As \(n \to \infty\), \(|a_N|^{1/n} \to 1\) because \(|a_N|\) is a fixed positive number, and \((n-N)/n \to 1\).
3. Q.E.D.
Step 2 holds for every \(\varepsilon > 0\), so \(\lim |a_n|^{1/n} = L\).