prime theorem of a convergent sequence, a

Theorem.

Suppose $(a_{n})$ is a positive real sequence that converges to $L$ . Thenthe sequence of arithmetic means $(b_{n})=(n^{-1}\\sum_{k=1}^{n}a_{k})$ 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 $n\\geq N_{0}$ 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 $N\\geq N_{0}$ for which $(N_{0}-1)M/N<\\varepsilon/2$ .

By the triangle inequality,

	$\\displaystyle\|b_{n}-L\|$	$\\displaystyle\\leq\\frac{1}{n}\\sum_{k=1}^{n}\|a_{k}-L\|$
		$\\displaystyle<\\frac{(N_{0}-1)M}{n}+\\frac{(n-N_{0}+1)\\varepsilon}{2n}$
		$\\displaystyle<\\varepsilon/2+\\varepsilon/2.$

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}}\\leq\\liminf\\sqrt[n]{d_{n}}\\leq\\limsup\\sqrt[n]{d_{n%}}\\leq\\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 $L$ .∎

References

1 Rudin, W., Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, New York, 1976.