Please help me to understand a proof that $a_n = frac{tan 1}{2} + frac{tan2 }{2^2} + dots + frac{tan n}{2^n}$...
$begingroup$
$textbf{Problem:}$ Show that $$a_n = frac{tan 1}{2} + frac{tan2 }{2^2} + dots + frac{tan n}{2^n}$$ is a Cauchy sequence.
This is the solution.
Here's my question: In the first line, we have $arctan$ or $tan$? $a_n$ is $tan$ or $arctan$? I mean the solution is wrong?
proof-explanation cauchy-sequences
edited Dec 28 '18 at 15:58
Blue
49.1k870156
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add a comment |
$begingroup$
$textbf{Problem:}$ Show that $$a_n = frac{tan 1}{2} + frac{tan2 }{2^2} + dots + frac{tan n}{2^n}$$ is a Cauchy sequence.
This is the solution.
Here's my question: In the first line, we have $arctan$ or $tan$? $a_n$ is $tan$ or $arctan$? I mean the solution is wrong?
proof-explanation cauchy-sequences
edited Dec 28 '18 at 15:58
Blue
49.1k870156
$endgroup$
add a comment |
$begingroup$
$textbf{Problem:}$ Show that $$a_n = frac{tan 1}{2} + frac{tan2 }{2^2} + dots + frac{tan n}{2^n}$$ is a Cauchy sequence.
This is the solution.
Here's my question: In the first line, we have $arctan$ or $tan$? $a_n$ is $tan$ or $arctan$? I mean the solution is wrong?
proof-explanation cauchy-sequences
edited Dec 28 '18 at 15:58
Blue
49.1k870156
$endgroup$
$textbf{Problem:}$ Show that $$a_n = frac{tan 1}{2} + frac{tan2 }{2^2} + dots + frac{tan n}{2^n}$$ is a Cauchy sequence.
This is the solution.
Here's my question: In the first line, we have $arctan$ or $tan$? $a_n$ is $tan$ or $arctan$? I mean the solution is wrong?
proof-explanation cauchy-sequences
proof-explanation cauchy-sequences
edited Dec 28 '18 at 15:58
Blue
49.1k870156
edited Dec 28 '18 at 15:58
Blue
49.1k870156
edited Dec 28 '18 at 15:58
Blue
49.1k870156
edited Dec 28 '18 at 15:58
Blue
49.1k870156
edited Dec 28 '18 at 15:58
Blue
49.1k870156
49.1k870156
asked Dec 28 '18 at 15:47
user462021
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2 Answers
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I'm quite sure that the question is wrong and that whoever asked it meant to write $arctan$ instead of $tan$.
answered Dec 28 '18 at 15:56
José Carlos SantosJosé Carlos Santos
168k23132236
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$begingroup$
The fact that $sum_{ngeq 1}frac{arctan n}{2^n}$ is convergent is fairly trivial, the fact that
$$sum_{ngeq 1}frac{tan n}{2^n} $$
is convergent... not so much. We may recall that $pi$ has a finite irrationality measure, hence
$$ left|frac{pi}{2}-frac{p}{q}right|leq frac{1}{q^{10}} $$
occurs for a finite number of $frac{p}{q}inmathbb{Q}$. In particular even if $left|tan nright|$ is large, since $n$ is close to an odd, integer multiple of $frac{pi}{2}$, it cannot be larger than, say, $n^{12}$, with a finite number of exceptions. Since $sum_{ngeq 1}frac{n^{12}}{2^n}$ is convergent, $sum_{ngeq 1}frac{tan n}{2^n}$ is convergent.
But dealing with $tan$ (in place of $arctan$) requires to invoke some fact from Diophantine approximation.
answered Dec 28 '18 at 21:46
Jack D'AurizioJack D'Aurizio
291k33284668
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
I'm quite sure that the question is wrong and that whoever asked it meant to write $arctan$ instead of $tan$.
answered Dec 28 '18 at 15:56
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
$begingroup$
I'm quite sure that the question is wrong and that whoever asked it meant to write $arctan$ instead of $tan$.
answered Dec 28 '18 at 15:56
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
$begingroup$
I'm quite sure that the question is wrong and that whoever asked it meant to write $arctan$ instead of $tan$.
answered Dec 28 '18 at 15:56
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
I'm quite sure that the question is wrong and that whoever asked it meant to write $arctan$ instead of $tan$.
answered Dec 28 '18 at 15:56
José Carlos SantosJosé Carlos Santos
168k23132236
answered Dec 28 '18 at 15:56
José Carlos SantosJosé Carlos Santos
168k23132236
answered Dec 28 '18 at 15:56
José Carlos SantosJosé Carlos Santos
168k23132236
answered Dec 28 '18 at 15:56
José Carlos SantosJosé Carlos Santos
168k23132236
168k23132236
add a comment |
add a comment |
$begingroup$
The fact that $sum_{ngeq 1}frac{arctan n}{2^n}$ is convergent is fairly trivial, the fact that
$$sum_{ngeq 1}frac{tan n}{2^n} $$
is convergent... not so much. We may recall that $pi$ has a finite irrationality measure, hence
$$ left|frac{pi}{2}-frac{p}{q}right|leq frac{1}{q^{10}} $$
occurs for a finite number of $frac{p}{q}inmathbb{Q}$. In particular even if $left|tan nright|$ is large, since $n$ is close to an odd, integer multiple of $frac{pi}{2}$, it cannot be larger than, say, $n^{12}$, with a finite number of exceptions. Since $sum_{ngeq 1}frac{n^{12}}{2^n}$ is convergent, $sum_{ngeq 1}frac{tan n}{2^n}$ is convergent.
But dealing with $tan$ (in place of $arctan$) requires to invoke some fact from Diophantine approximation.
answered Dec 28 '18 at 21:46
Jack D'AurizioJack D'Aurizio
291k33284668
$endgroup$
add a comment |
$begingroup$
The fact that $sum_{ngeq 1}frac{arctan n}{2^n}$ is convergent is fairly trivial, the fact that
$$sum_{ngeq 1}frac{tan n}{2^n} $$
is convergent... not so much. We may recall that $pi$ has a finite irrationality measure, hence
$$ left|frac{pi}{2}-frac{p}{q}right|leq frac{1}{q^{10}} $$
occurs for a finite number of $frac{p}{q}inmathbb{Q}$. In particular even if $left|tan nright|$ is large, since $n$ is close to an odd, integer multiple of $frac{pi}{2}$, it cannot be larger than, say, $n^{12}$, with a finite number of exceptions. Since $sum_{ngeq 1}frac{n^{12}}{2^n}$ is convergent, $sum_{ngeq 1}frac{tan n}{2^n}$ is convergent.
But dealing with $tan$ (in place of $arctan$) requires to invoke some fact from Diophantine approximation.
answered Dec 28 '18 at 21:46
Jack D'AurizioJack D'Aurizio
291k33284668
$endgroup$
add a comment |
$begingroup$
The fact that $sum_{ngeq 1}frac{arctan n}{2^n}$ is convergent is fairly trivial, the fact that
$$sum_{ngeq 1}frac{tan n}{2^n} $$
is convergent... not so much. We may recall that $pi$ has a finite irrationality measure, hence
$$ left|frac{pi}{2}-frac{p}{q}right|leq frac{1}{q^{10}} $$
occurs for a finite number of $frac{p}{q}inmathbb{Q}$. In particular even if $left|tan nright|$ is large, since $n$ is close to an odd, integer multiple of $frac{pi}{2}$, it cannot be larger than, say, $n^{12}$, with a finite number of exceptions. Since $sum_{ngeq 1}frac{n^{12}}{2^n}$ is convergent, $sum_{ngeq 1}frac{tan n}{2^n}$ is convergent.
But dealing with $tan$ (in place of $arctan$) requires to invoke some fact from Diophantine approximation.
answered Dec 28 '18 at 21:46
Jack D'AurizioJack D'Aurizio
291k33284668
$endgroup$
The fact that $sum_{ngeq 1}frac{arctan n}{2^n}$ is convergent is fairly trivial, the fact that
$$sum_{ngeq 1}frac{tan n}{2^n} $$
is convergent... not so much. We may recall that $pi$ has a finite irrationality measure, hence
$$ left|frac{pi}{2}-frac{p}{q}right|leq frac{1}{q^{10}} $$
occurs for a finite number of $frac{p}{q}inmathbb{Q}$. In particular even if $left|tan nright|$ is large, since $n$ is close to an odd, integer multiple of $frac{pi}{2}$, it cannot be larger than, say, $n^{12}$, with a finite number of exceptions. Since $sum_{ngeq 1}frac{n^{12}}{2^n}$ is convergent, $sum_{ngeq 1}frac{tan n}{2^n}$ is convergent.
But dealing with $tan$ (in place of $arctan$) requires to invoke some fact from Diophantine approximation.
answered Dec 28 '18 at 21:46
Jack D'AurizioJack D'Aurizio
291k33284668
answered Dec 28 '18 at 21:46
Jack D'AurizioJack D'Aurizio
291k33284668
answered Dec 28 '18 at 21:46
Jack D'AurizioJack D'Aurizio
291k33284668
answered Dec 28 '18 at 21:46
Jack D'AurizioJack D'Aurizio
291k33284668
291k33284668
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