prime theorem of a convergent sequence, aAPrimeTheoremOfAConvergentSequence2004-11-18 21:04:032006-11-14 11:53:01Theorem% this is the default PlanetMath preamble. as your knowledge
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\begin{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 $L$.
\end{theorem*}
\begin{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 $|a_n-L|<\varepsilon/2$. 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 $N\ge N_0$ for which $(N_0-1)M/N<\varepsilon/2$.
By the triangle inequality,
\begin{align*}
|b_n-L|
&\le\frac{1}{n}\sum_{k=1}^n|a_k-L| \\
&<\frac{(N_0-1)M}{n}+\frac{(n-N_0+1)\varepsilon}{2n} \\
&<\varepsilon/2+\varepsilon/2.
\end{align*}
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
\[
\liminf \frac{d_{n+1}}{d_n} \le
\liminf \sqrt[n]{d_n} \le
\limsup \sqrt[n]{d_n} \le
\limsup \frac{d_{n+1}}{d_n},
\]
a proof of which can be found in~\cite{Ru}. 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 $L$.
\end{proof}
\begin{thebibliography}{1}
\bibitem{Ru}
Rudin, W., \emph{Principles of Mathematical Analysis}, 3rd ed., McGraw-Hill, New York, 1976.
\end{thebibliography}