# 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| 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

Title prime theorem of a convergent sequence, a PrimeTheoremOfAConvergentSequenceA 2013-03-22 14:49:45 2013-03-22 14:49:45 georgiosl (7242) georgiosl (7242) 24 georgiosl (7242) Theorem msc 40-00 ArithmeticMean GeometricMean