| Version 14 |
Version 13 |
| Statement: If $a_n$ is a convergent sequence $(a_n \in \R)$, 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 $(a_n \in \R)$, 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: |
| We have $\lim_{n \to +\infty} a_n =a$, and the sequence of geometric means is defined to be the $b_n= \sqrt[n]{a_1\dots a_n}$. |
We have $\lim_{n \to +\infty} a_n =a$, and the sequence of geometric means is defined to be the $b_n= \sqrt[n]{a_1\dots a_n}$. |
| We set $c_n = a_1\cdots a_n$, then |
We set $c_n = a_1\cdots a_n$, then |
| \[ \lim_{n \to +\infty} \frac{c_{n+1}}{c_n}= a_{n+1}.\] |
\[ \lim_{n \to +\infty} \frac{c_{n+1}}{c_n}= a_{n+1}.\] |
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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: |
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\[ \liminf \frac{t_{n+1}}{t_n}\leq \liminf \sqrt[n]{t_n} \leq \limsup \sqrt[n]{t_n} \leq \limsup \frac{t_{n+1}}{t_n}. \]
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\[ \liminf \frac{t_{n+1}}{t_n}\leq \liminf \sqrt[n]{t_n} \leq limsup \sqrt[n]{t_n} \leq \limsup \frac{t_{n+1}}{t_n}. \]
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| 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. |
| In case where \[ \lim_{n \to +\infty} a(n) =0\]. |
In case where \[ \lim_{n \to +\infty} a(n) =0\]. |
| For a given $\epsilon\geq 0$ we find $k \in \aleph_0$ satisfying |
For a given $\epsilon\geq 0$ we find $k \in \aleph_0$ satisfying |
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\[\forall n \geq \kappa -\epsilon \leq a \leq \epsilon \Rightarrow | \frac{1){n-k} | a_{k+1} a_{k+2}\dots a_n | \leq \frac{1}{2} \epsilon \]
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\[\forall n \geq \kappa -\epsilon \leq a \leq \epsilon \Rightarrow | \frac{1){n-k} | |a_{k+1} a_{k+2}\dots a_n | \ll \frac{1}{2} \epsilon \]
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| (by the definition of convergence). |
(by the definition of convergence). |
| Also we have for the concrete $k$, we have choiced and for a suitable $\nu$ that: |
Also we have for the concrete $k$, we have choiced and for a suitable $\nu$ that: |
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\[ \|frac {1} {\nu} |a_1+...+a_\nu|\]
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\[ \|frac {1} {\nu} |$a_1+...+a_\nu|\]
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| Now we apply triangular inequality so: |
Now we apply triangular inequality so: |
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\[ \|frac {1} {\nu} |a_1+...+a_\nu)| \leq frac {1} {\nu} |a_1+...+a_k|+ frac {1} {\nu} |a_1+....+a_\nu)\]
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\[ \|frac {1} {\nu} |$a_1+...+a\nu)$| \leq frac {1} {\nu} |$a(1)+...+a(k)$|+ frac {1} {\nu} |$a_1+....+a\nu)\]
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| 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 $a_n-a$ |
