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Version 20 |
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| \begin{theorem*} |
\begin{theorem*} |
| Suppose $(a_n)$ is a positive real sequence that converges to $L$. Then |
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 $L$. |
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*} |
\end{theorem*} |
|
|
| \begin{proof} |
\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
|
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 $|a_n-L|<\varepsilon/3$. Since $(a_n)$ converges to a finite value, there is a finite $M$ such that
|
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$|a_n-L|<M$ for all $n$. Thus we can select a positive integer $N\ge N_0$ for which $(N_0-1)M/N<\varepsilon/2$.
|
$|a_n-L|<M$ for all $n$. Thus we can select a positive integer $N\ge N_0$ for which $(N-1)M/N<\varepsilon/3$.
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| By the triangle inequality, |
By the triangle inequality, |
| \begin{align*} |
\begin{align*} |
| |b_n-L| |
|b_n-L| |
| &\le\frac{1}{n}\sum_{k=1}^n|a_k-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} \\
|
&<\frac{(N-1)M}{n}+\frac{(n-N+1)\varepsilon}{3n} \\
|
|
&<\varepsilon/2+\varepsilon/2.
|
&<\varepsilon/3+2\varepsilon/3.
|
| \end{align*} |
\end{align*} |
| Hence $(b_n)$ converges to $L$. |
Hence $(b_n)$ converges to $L$. |
|
|
| To show that $(c_n)$ converges to $L$, we first define the sequence $(d_n)$ |
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 |
by $d_n=c_n^n=a_1\cdots a_n$. Since $d_n$ is a positive real sequence, we have that |
| \[ |
\[ |
| \liminf \frac{d_{n+1}}{d_n} \le |
\liminf \frac{d_{n+1}}{d_n} \le |
| \liminf \sqrt[n]{d_n} \le |
\liminf \sqrt[n]{d_n} \le |
| \limsup \sqrt[n]{d_n} \le |
\limsup \sqrt[n]{d_n} \le |
| \limsup \frac{d_{n+1}}{d_n}, |
\limsup \frac{d_{n+1}}{d_n}, |
| \] |
\] |
| 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 $L$. |
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} |
\end{proof} |
|
|
| \begin{thebibliography}{1} |
\begin{thebibliography}{1} |
| \bibitem{Ru} |
\bibitem{Ru} |
| Rudin, W., \emph{Principles of Mathematical Analysis}, 3rd ed., McGraw-Hill, New York, 1976. |
Rudin, W., \emph{Principles of Mathematical Analysis}, 3rd ed., McGraw-Hill, New York, 1976. |
| \end{thebibliography} |
\end{thebibliography} |
