# Elementary way to show $\lim_{n \rightarrow \infty} \sqrt[n]{a_n} = \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$?

Let $a_n \gt 0$ for $n \in \mathbb{N}$. The convergence radius of the series $\sum_{n=0}^\infty a_n z^n$ is $\frac{1}{q}$ with $q = \lim_{n \rightarrow \infty} \sqrt[n]{a_n}$ or $q = \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$, if these limits exist. Therefore the $\lim$s must be identical.

However I was wondering whether there exists a more elementary way to show this identity? (Or generally any other way?)

• "Therefore the limsups must be identical." No. Find a counterexample. (And the radius of convergence of the series $\sum a_nz^n$ with positive terms is not always the inverse of $\limsup_{n \rightarrow \infty}a_{n+1}/a_n$.)
– Did
Jan 20, 2015 at 8:34
• What are $a_n$, functions or numbers? Jan 20, 2015 at 8:36
• @JankoBracic: Real numbers.
– Keba
Jan 20, 2015 at 8:44
• @Did: Oh, your‘re right. Well, then I should assume the existence of both limits. I‘ll edit the question.
– Keba
Jan 20, 2015 at 8:46
• @Keba But what is then the convergence radius of $\sum a_n$? Jan 20, 2015 at 8:46

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 $$\frac{a_{n+1}}{a_n} < S + \epsilon \quad \text{ for } n \ge N \, .$$ $$\implies a_n < a_N \, (S + \epsilon)^{n - N} \quad \text{ for } n \ge N$$ $$\implies \sqrt[n]{\mathstrut a_n} < (S + \epsilon) \sqrt[n]{ a_N (S + \epsilon)^{-N}}$$ $$\implies \limsup_{n \to \infty} \sqrt[n]{\mathstrut a_n} \le S + \epsilon \, .$$
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.