I can't figure out how to prove this inequality, $a_n \gt 0$ is a bounded sequence.
$$\liminf_{n\to \infty} a_n\le\liminf_{n\to \infty} \sqrt[n]{a_1a_2...a_n} \le \limsup_{n\to \infty} \sqrt[n]{a_1a_2...a_n}\le \limsup_{n\to \infty}a_n$$
I hope I'm correct in refering to $\sqrt[n]{a_1a_2...a_n}$ as the geometric mean.
Also while I'm not sure if it helps I have that: $$\sqrt[n]{a_1a_2...a_n}=\exp{(\frac{\log{a_1}+\log{a_2}+...+\log{a_n}}{n})}$$
I have proven it if $a_n$ is convergent (pretty trivial). Any help would be greatly appreciated.
The middle inequality follows from $\liminf_{n\to\infty}t_n\leq\limsup_{n\to\infty}t_n$ for any sequence $t_n$. The first and the third one have similar proofs, so I'll show only the third one. Suppose $\limsup_{n\to\infty}a_n=A\geq0$ (note $a_n>0$ and $a_n$ is bounded). Let $\varepsilon>0$. There is $N\in\mathbb{N}$ such that $a_n<A+\varepsilon$ when $n>N$. Then $$\sqrt[n]{a_1a_2\cdots a_n}<\sqrt[n]{a_1\cdots a_N\left(A+\varepsilon\right)^{n-N}}=\sqrt[n]{\frac{a_1a_2\cdots a_N}{\left(A+\varepsilon\right)^N}}(A+\varepsilon) $$ Since $$\lim_{n\to\infty}\sqrt[n]{\frac{a_1a_2\cdots a_N}{\left(A+\varepsilon\right)^N}}=1 $$ there is $M\in\mathbb{N}$ such that $$\sqrt[n]{\frac{a_1a_2\cdots a_N}{\left(A+\varepsilon\right)^N}}<1+\eta $$ when $n>M$, and we've chosen $0<\eta<\frac{\varepsilon}{A+\varepsilon}$. So, for $n>\max(N,M)$, we have $$\sqrt[n]{a_1a_2\cdots a_n}<(1+\eta)(A+\varepsilon)<A+2\varepsilon $$ $\varepsilon$ is arbitrary, which implies $\limsup_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}\leq A $, as was claimed.