<s>[INST] show that $\lim\limits_{n\to\infty} \frac{a_1+\cdots + a_n}n = 1$ 
Let $(a_n)$ be a bounded sequence of positive real numbers such that $\lim\limits_{n\to\infty} \sqrt[n]{a_1\cdots a_n} = 1.$ Prove that $\lim\limits_{n\to\infty} \dfrac{a_1+\cdots + a_n}n = 1.$

By the AM-GM inequality, $ \sqrt[n]{a_1\cdots a_n} \leq \dfrac{a_1+\cdots + a_n}n $ for all $n$ so taking limits on both sides yields $\lim\limits_{n\to\infty} \frac{a_1+\cdots + a_n}n \ge \lim\limits_{n\to\infty} \sqrt[n]{a_1\cdots a_n} = 1.$ Thus it suffices to show that $\lim\limits_{n\to\infty} \dfrac{a_1+\cdots + a_n}n \leq 1.$ Also, if a sequence of real numbers $(a_n)$ satisfies $\lim\limits_{n\to\infty} a_n = L,$ then $\lim\limits_{n\to\infty} \dfrac{a_1+\cdots + a_n}n = L,$ so if we could show that $\lim\limits_{n\to\infty} a_n = 1,$ we would be done. Suppose without loss of generality that $L = \lim\limits_{n\to\infty} a_n > 1$ (the case where $L < 1$ is similar). Let $N > 1$ be such that $a_n > \frac{1+L}2$ for all $n\ge N.$ Then for $n \ge N, \sqrt[n]{a_1\cdots a_n} = \sqrt[n]{a_1\cdots a_{N-1}}\sqrt[n]{a_N \cdots a_n} > R_n \sqrt[n]{(\frac{1+L}2)^{n-N+1}}$, where $R_n = \sqrt[n]{a_1\cdots a_{N-1}}.$ Hence taking the limit of both sides (for $n\ge N$) yields $\lim\limits_{n\to\infty} \sqrt[n]{a_1\cdots a_n}\ge \lim\limits_{n\to\infty} R_n (\dfrac{1+L}2)^{(n-N+1)/n} = \frac{1+L}2 > 1,$ a contradiction. Hence $\lim\limits_{n\to\infty} a_n = 1,$ so the result follows.

Is there something missing, or is the above proof correct?

 [/INST] This is false. Consider the sequence
$$a_{j} = \begin{cases}2 && \text{if }j\text{ is odd}\\
\frac{1}{2} && \text{if }j \text{ is even} \end{cases}$$
we have for all $n$
$$1 \leq (a_{1}...a_{n})^{\frac{1}{n}} \leq 2^{\frac{1}{n}}$$
Thus
$$\lim_{n \rightarrow \infty}(a_{1}...a_{n})^{\frac{1}{n}} = 1$$
But
$$\lim_{n \rightarrow \infty}\frac{\sum_{j=1}^{2n}a_{j}}{2n} = \lim_{n \rightarrow \infty}\frac{2n+\frac{1}{2}n}{2n} = \frac{5}{4}.$$
</s>