If $\{a_n\}$ converges to $A$, then $\{(a_1\cdots a_n)^{1/n}\}$ converges to $A$

Prove that this sequence converges. I can't do it.

Let $\{a_n\}$ be a sequence of positive real numbers that converges to a number $A$. Prove that $\{(a_1\cdots a_n)^{1/n}\}$ converges to $A$.

2 Answers

Since $a_n$ converges to $A$, and all numbers are positive, it follows that $\log(a_n)$ converges to $\log(A)$. By Cesaro's averaging theorem:

$$\frac{\sum_{i=1}^n \log(a_i)}{n}\rightarrow \log(A).$$

Exponentiating both sides gives the desired result.

Let $$x_n=\ln(a_1a_2\ldots a_n)^{1/n}=\frac1n\sum_{i=1}^n\ln a_i.$$ Since $\lim_na_n=A$, then $\lim_n\ln a_n=\ln A$, and therefore the sequence $(x_n)$ is convergent, with $\lim_nx_n=\ln A$. It follows that $$\lim_n(a_1a_2\ldots a_n)^{1/n}=\lim_ne^{x_n}=e^{\ln A}=A.$$

• why is $x_n$ convergent? The $n$-th term is $a_n/n$.. I'd say that since the terms are (defintely) bigger then $(\ln A - 1)/n$, that series diverges.. not sure though – Ant Jul 7 '15 at 21:29
• @Ant This is a well-known result. Check en.wikipedia.org/wiki/Ces%C3%A0ro_mean – Mercy King Jul 7 '15 at 21:32
• oh, I see. Didn't know about cesaro mean before.. Should check out the proof :) – Ant Jul 7 '15 at 21:37