Theorem   Suppose $(a_n)$ is a positive real sequence that converges to $L$ . Then the sequence of arithmetic means $(b_n)=(n^{-1}\sum_{k=1}^na_k)$ and the sequence of geometric means $(c_n)=(\sqrt[n]{a_1\cdots a_n})$ also converge to $L$ .

Proof. We first show that $(b_n)$ converges to $L$ . Let $\varepsilon>0$ . Select a positive integer $N_0$ such that $n\ge N_0$ implies $|a_n-L|<\varepsilon/2$ . Since $(a_n)$ converges to a finite value, there is a finite $M$ such that $|a_n-L|<M$ for all $n$ . Thus we can select a positive integer $N\ge N_0$ for which $(N_0-1)M/N<\varepsilon/2$ .

By the triangle inequality,

Hence $(b_n)$ converges to $L$ .

To show that $(c_n)$ converges to $L$ , we first define the sequence $(d_n)$ by $d_n=c_n^n=a_1\cdots a_n$ . Since $d_n$ is a positive real sequence, we have that$$ \liminf \frac{d_{n+1}}{d_n} \le \liminf \sqrt[n]{d_n} \le \limsup \sqrt[n]{d_n} \le \limsup \frac{d_{n+1}}{d_n},$$ a proof of which can be found in [1]. But $d_{n+1}/d_n=a_{n+1}$ , which by assumption converges to $L$ . Hence $\sqrt[n]{d_n}=c_n$ must also converge to $L$ .