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Version 17 |
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| Statement: If $(a_n)$ is a convergent sequence of positive reals, then its sequence of arithmetic means $(t_n)$ and its sequence of geometric means $(b_n)$ converge to the same limit as $(a_n)$. |
Statement: If $(a_n)$ is a convergent sequence of positive reals, then its sequence of arithmetic means $(t_n)$ and its sequence of geometric means $(b_n)$ converge to the same limit as $(a_n)$. |
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| Proof: |
Proof: |
| Put $a=\lim_{n \to \infty} a_n$. The sequence of geometric means is defined by $b_n= \sqrt[n]{a_1\dots a_n}$. |
Put $a=\lim_{n \to \infty} a_n$. The sequence of geometric means is defined by $b_n= \sqrt[n]{a_1\dots a_n}$. |
| We set $c_n = a_1\cdots a_n$, then |
We set $c_n = a_1\cdots a_n$, then |
| \[ \frac{c_{n+1}}{c_n}= a_{n+1}.\] |
\[ \frac{c_{n+1}}{c_n}= a_{n+1}.\] |
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| Corollary in Analysis (Chapter Sequences): For any sequence of positive real numbers the following hold: |
Corollary in Analysis (Chapter Sequences): For any sequence of positive real numbers the following hold: |
| \[ \liminf \frac{t_{n+1}}{t_n}\leq \liminf \sqrt[n]{t_n} \leq \limsup \sqrt[n]{t_n} \leq \limsup \frac{t_{n+1}}{t_n}. \] |
\[ \liminf \frac{t_{n+1}}{t_n}\leq \liminf \sqrt[n]{t_n} \leq \limsup \sqrt[n]{t_n} \leq \limsup \frac{t_{n+1}}{t_n}. \] |
| From this corollary we have that \[ \lim_{n \to \infty}\sqrt[n]{a_n}= a.\] |
From this corollary we have that \[ \lim_{n \to \infty}\sqrt[n]{a_n}= a.\] |
| So we show that the geometric sequence $b_n$ and $a_n$ have the same limit. |
So we show that the geometric sequence $b_n$ and $a_n$ have the same limit. |
| Also we have to show that this of arithmetic sequence is the same too. |
Also we have to show that this of arithmetic sequence is the same too. |
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In case where \[ \lim_{n \to \infty} a_n =0.\]
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In case where \[ \lim_{n \to +\infty} a_n =0.\]
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| For a given $\epsilon\geq 0$ we find $k \in \N$ satisfying |
For a given $\epsilon\geq 0$ we find $k \in \N$ satisfying |
| \[\forall n \geq k -\epsilon \leq a \leq \epsilon \Rightarrow | |
\[\forall n \geq k -\epsilon \leq a \leq \epsilon \Rightarrow | |
| \frac{1}{n-k} | a_{k+1} a_{k+2}\dots a_n | \leq \frac{1}{2} \epsilon \] |
\frac{1}{n-k} | a_{k+1} a_{k+2}\dots a_n | \leq \frac{1}{2} \epsilon \] |
| (by the definition of convergence). |
(by the definition of convergence). |
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Also, for the $k$ we have chosen there is a $\nu$ such that:
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Also we have for the concrete $k$, we have choiced and for a suitable $\nu$ that:
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| \[ \frac {1} {\nu} |a_1+...+a_\nu|\] |
\[ \frac {1} {\nu} |a_1+...+a_\nu|\] |
| Now we apply triangular inequality so: |
Now we apply triangular inequality so: |
| \[ \frac {1} {\nu} |a_1+...+a_\nu)| \leq \frac{1}{\nu} |a_1+...+a_k|+ \frac{1}{\nu} |a_1+....+a_\nu)\] |
\[ \frac {1} {\nu} |a_1+...+a_\nu)| \leq \frac{1}{\nu} |a_1+...+a_k|+ \frac{1}{\nu} |a_1+....+a_\nu)\] |
| Q.E.D |
Q.E.D |
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| In the other case where \[ \lim_{n \to +\infty} a_n =a \not= 0\] we follow the same steps but this time we use $a_n-a$ |
In the other case where \[ \lim_{n \to +\infty} a_n =a \not= 0\] we follow the same steps but this time we use $a_n-a$ |
