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Suppose $a_n\geq 0$ for all $n$ and $$\lim_{n\to \infty }\frac{a_{n+1}}{a_n}=\rho.$$

How can I show that $\lim_{n\to \infty }\sqrt[n]{a_n}=\rho$ ?

Atempts $$|\sqrt[n]{a_n}-\rho|=\frac{|a_n-\rho|}{a_n^{n-1}+\rho a_n^{n-2}+...+\rho^{n-1}},$$ but I'm not so sure why the RHS goes to 0 when $n\to \infty $.

Bruce
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  • Also: https://math.stackexchange.com/q/69386/42969, https://math.stackexchange.com/a/76800/42969 – Martin R Oct 14 '20 at 19:11
  • $\ln(a_{n+1})-\ln(a_n)\underset{n\rightarrow +\infty}{\longrightarrow}\ln\rho$ thus using Cesaro's theorem, $\frac{\ln a_n}{n}\underset{n\rightarrow +\infty}{\longrightarrow}\ln\rho$ and $\sqrt[n]{a_n}\underset{n\rightarrow +\infty}{\longrightarrow}\rho$. – Tuvasbien Oct 14 '20 at 19:12
  • What do you mean by Cesaro ? The only Cesaro theorem I know is $x_n\to \ell$ implies $\frac{1}{n}\sum_{k=1}^n x_k\to \ell$. You use this one ? – Bruce Oct 14 '20 at 19:17

2 Answers2

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Hint : Take the $\ln$ and apply Cesaro.

TheSilverDoe
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Let $\varepsilon >0$. There is $N$ s.t. $$(\rho-\varepsilon)|a_{n-1}| \leq |a_{n}|\leq (1+\varepsilon )|a_{n-1}|, $$ for all $n>N$. And thus, if $n>N$, $$(\rho-\varepsilon )^{n-N}|a_{N}|\leq |a_n|\leq (\rho+\varepsilon )^{n-N}|a_N|,$$ i.e. $$(\rho-\varepsilon )^{\frac{n-N}{n}}\leq \sqrt[n]{a_n}\leq (\rho+\varepsilon )^{\frac{n-N}{n}}.$$

Therefore $$\limsup_{n\to \infty }\sqrt[n]{a_n}\leq \rho +\varepsilon \quad \text{and}\quad \liminf_{n\to \infty }\sqrt[n]{a_n}\geq \rho-\varepsilon .$$

Since the result hold for all $\varepsilon >0$, the claim follow.

Surb
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  • Why do you use limsup and liminf and not the limit directly ? – Bruce Oct 14 '20 at 19:18
  • Because, there is no reason that $\lim_{n\to \infty }\sqrt[n]{a_n}$ exist a priori. – Surb Oct 14 '20 at 19:20
  • Sorry, I don't get your point, you prove that $\rho-\varepsilon \leq \lim_{n\to \infty }\sqrt[n]{a_n}\leq \rho+\varepsilon $, and thus, since this hold for all $\varepsilon $, the limit exist, no ? – Bruce Oct 14 '20 at 19:22
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    Well, you can't do that's way. $a_n\leq b_n\leq c_n$ for all $n$ and $a_n\to a$ and $b_n\to b$, doesn't implies that $\lim_{n\to \infty }b_n$ exist (here $a$ and $b$ are different, so squeeze theorem doesn't apply if you were thinking about that). A simple example $-1\leq (-1)^n\leq 1$.... – Surb Oct 14 '20 at 19:25