prime theorem of a convergent sequence, a

Theorem.

Suppose $(a_{n})$ is a positive real sequence that converges to $L$ . Then the sequence of arithmetic means $(b_{n})=(n^{-1}\sum_{k=1}^{n}a_{k})$ 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.

Title	prime theorem of a convergent sequence, a
Canonical name	PrimeTheoremOfAConvergentSequenceA
Date of creation	2013-03-22 14:49:45
Last modified on	2013-03-22 14:49:45
Owner	georgiosl (7242)
Last modified by	georgiosl (7242)
Numerical id	24
Author	georgiosl (7242)
Entry type	Theorem
Classification	msc 40-00
Related topic	ArithmeticMean
Related topic	GeometricMean