# What about the convergence of the geometric mean sequence of the terms of a given convergent sequence?

Let $(a_n)$ be a sequence of positive real numbers such that $$\lim_{n\to} a_n = a.$$ Let $b_n \colon= \sqrt[n]{a_1 \cdot a_2 \cdot a_3 \cdots a_n}$ and $c_n \colon= a_n^{a_n}$. Then what can we say about $$\lim_{n\to\infty} b_n$$ and $$\lim_{n\to\infty} c_n?$$

What if $(a_n)$ is a sequence of complex numbers instead?

$$\log b_n=\frac{\log a_1+\cdots+\log a_n}{n}.$$ By Stolz's Theorem and some easy computation, $$\lim_{n\rightarrow\infty}b_n=a.$$ $$\lim_{n\rightarrow\infty}c_n=a^a.$$ Here $0^0$ is regarded to be $1$.
• Morris, how do we know that if $a_n \to a$ as $n \to \infty$, then $\log a_n \to log a$? How to rigorously prove this? Commented Oct 31, 2014 at 15:29
• @Saaqib Mahmuud Without loss of generality, assume $a>0$. When $n$ is sufficiently large, we have $\frac{a}{2}<a_n<2a$. By the Lagrange mean value theorem, $|\log a_n-\log a|=|\frac{1}{t_n}(a_n-a)|\leq\frac{2}{a}|a_n-a|\rightarrow 0$. Commented Oct 31, 2014 at 15:57
• But I would like an $\epsilon$ argument! Commented Nov 2, 2014 at 8:52