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a prime theorem of a convergent sequence (Theorem)
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 $ \vert a_n-L\vert<\varepsilon/2$. Since $ (a_n)$ converges to a finite value, there is a finite $ M$ such that $ \vert a_n-L\vert<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,

$\displaystyle \vert b_n-L\vert$ $\displaystyle \le\frac{1}{n}\sum_{k=1}^n\vert a_k-L\vert$    
  $\displaystyle <\frac{(N_0-1)M}{n}+\frac{(n-N_0+1)\varepsilon}{2n}$    
  $\displaystyle <\varepsilon/2+\varepsilon/2.$    

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

$\displaystyle \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$. $ \qedsymbol$

Bibliography

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



"a prime theorem of a convergent sequence" is owned by georgiosl. [ full author list (3) | owner history (5) ]
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See Also: arithmetic mean, geometric mean
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Cross-references: proof, triangle inequality, finite, implies, integer, geometric means, arithmetic means, converges, sequence, real, positive

This is version 21 of a prime theorem of a convergent sequence, born on 2004-11-18, modified 2006-11-14.
Object id is 6494, canonical name is APrimeTheoremOfAConvergentSequence.
Accessed 2642 times total.

Classification:
AMS MSC: 40-00 (Sequences, series, summability :: General reference works )
Pending Errata and Addenda
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error by rgreenlee on 2006-10-17 19:53:37
i see the error too. how do i file a correction? do i simply write up some LaTeX? i had this problem on an analysis test and have a nice proof.
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Choosing N by radarnasty on 2006-10-17 15:19:52
There is a problem with the first part of this proof because by composition M could be an arbitrarily large finite number and since the integer N is going to be large (N-1)M/N is going to approach M which is not necessarily going to be less than epsilon/3.
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