Trying to prove $e$'s irrationality
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3
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Knowing that $limlimits_{xto 0} $$frac{sin(x)}{x}$$= 1$ , $frac{1}{n+1}<n!r_n<frac{1}{n}$, where $r_n=e- sum _{ k=0 }^{ n}{ frac { 1 }{k!}} $
By studying $limlimits_{ntoinfty} nsin(2πn!r_n)$ I have to show that :
$$limlimits_{ntoinfty} nsin(2πn!e)= 2π,$$
and then prove that $e$ is irrational ?
limits exponential-function irrational-numbers
asked Nov 18 at 11:58
Lamethyste
424
add a comment |
up vote
3
down vote
favorite
Knowing that $limlimits_{xto 0} $$frac{sin(x)}{x}$$= 1$ , $frac{1}{n+1}<n!r_n<frac{1}{n}$, where $r_n=e- sum _{ k=0 }^{ n}{ frac { 1 }{k!}} $
By studying $limlimits_{ntoinfty} nsin(2πn!r_n)$ I have to show that :
$$limlimits_{ntoinfty} nsin(2πn!e)= 2π,$$
and then prove that $e$ is irrational ?
limits exponential-function irrational-numbers
asked Nov 18 at 11:58
Lamethyste
424
Well, ok. The first limit seems manageable. What are your thoughts?
– Yuriy S
Nov 18 at 12:17
I tried by composition because $limlimits_{ntoinfty} 2πn!r_n = 0$ and I showed $$limlimits_{ntoinfty} nsin(2πn!r_n)= 2π$$ but i don't think that's correct
– Lamethyste
Nov 18 at 12:26
Seems fine to me. But not sure about the other limit
– Yuriy S
Nov 18 at 12:48
Shouldn't the sum in the definition of $r_n$ start at $k=0$ instead of $k=1$?
– Barry Cipra
Nov 18 at 12:50
You are right, my bad!
– Lamethyste
Nov 18 at 12:55
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Knowing that $limlimits_{xto 0} $$frac{sin(x)}{x}$$= 1$ , $frac{1}{n+1}<n!r_n<frac{1}{n}$, where $r_n=e- sum _{ k=0 }^{ n}{ frac { 1 }{k!}} $
By studying $limlimits_{ntoinfty} nsin(2πn!r_n)$ I have to show that :
$$limlimits_{ntoinfty} nsin(2πn!e)= 2π,$$
and then prove that $e$ is irrational ?
limits exponential-function irrational-numbers
asked Nov 18 at 11:58
Lamethyste
424
Knowing that $limlimits_{xto 0} $$frac{sin(x)}{x}$$= 1$ , $frac{1}{n+1}<n!r_n<frac{1}{n}$, where $r_n=e- sum _{ k=0 }^{ n}{ frac { 1 }{k!}} $
By studying $limlimits_{ntoinfty} nsin(2πn!r_n)$ I have to show that :
$$limlimits_{ntoinfty} nsin(2πn!e)= 2π,$$
and then prove that $e$ is irrational ?
limits exponential-function irrational-numbers
limits exponential-function irrational-numbers
asked Nov 18 at 11:58
Lamethyste
424
asked Nov 18 at 11:58
Lamethyste
424
edited Nov 18 at 12:55
asked Nov 18 at 11:58
Lamethyste
424
asked Nov 18 at 11:58
Lamethyste
424
asked Nov 18 at 11:58
Lamethyste
424
424
Well, ok. The first limit seems manageable. What are your thoughts?
– Yuriy S
Nov 18 at 12:17
I tried by composition because $limlimits_{ntoinfty} 2πn!r_n = 0$ and I showed $$limlimits_{ntoinfty} nsin(2πn!r_n)= 2π$$ but i don't think that's correct
– Lamethyste
Nov 18 at 12:26
Seems fine to me. But not sure about the other limit
– Yuriy S
Nov 18 at 12:48
Shouldn't the sum in the definition of $r_n$ start at $k=0$ instead of $k=1$?
– Barry Cipra
Nov 18 at 12:50
You are right, my bad!
– Lamethyste
Nov 18 at 12:55
add a comment |
Well, ok. The first limit seems manageable. What are your thoughts?
– Yuriy S
Nov 18 at 12:17
I tried by composition because $limlimits_{ntoinfty} 2πn!r_n = 0$ and I showed $$limlimits_{ntoinfty} nsin(2πn!r_n)= 2π$$ but i don't think that's correct
– Lamethyste
Nov 18 at 12:26
Seems fine to me. But not sure about the other limit
– Yuriy S
Nov 18 at 12:48
Shouldn't the sum in the definition of $r_n$ start at $k=0$ instead of $k=1$?
– Barry Cipra
Nov 18 at 12:50
You are right, my bad!
– Lamethyste
Nov 18 at 12:55
Well, ok. The first limit seems manageable. What are your thoughts?
– Yuriy S
Nov 18 at 12:17
Well, ok. The first limit seems manageable. What are your thoughts?
– Yuriy S
Nov 18 at 12:17
I tried by composition because $limlimits_{ntoinfty} 2πn!r_n = 0$ and I showed $$limlimits_{ntoinfty} nsin(2πn!r_n)= 2π$$ but i don't think that's correct
– Lamethyste
Nov 18 at 12:26
I tried by composition because $limlimits_{ntoinfty} 2πn!r_n = 0$ and I showed $$limlimits_{ntoinfty} nsin(2πn!r_n)= 2π$$ but i don't think that's correct
– Lamethyste
Nov 18 at 12:26
Seems fine to me. But not sure about the other limit
– Yuriy S
Nov 18 at 12:48
Seems fine to me. But not sure about the other limit
– Yuriy S
Nov 18 at 12:48
Shouldn't the sum in the definition of $r_n$ start at $k=0$ instead of $k=1$?
– Barry Cipra
Nov 18 at 12:50
Shouldn't the sum in the definition of $r_n$ start at $k=0$ instead of $k=1$?
– Barry Cipra
Nov 18 at 12:50
You are right, my bad!
– Lamethyste
Nov 18 at 12:55
You are right, my bad!
– Lamethyste
Nov 18 at 12:55
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Let $I_n=sumlimits_{k=0}^ndfrac{n!}{k!}in Bbb Z$, then
$$nsin(2pi n! e)=nsin(2pi I_n+2pi n!r_n)=nsin(2pi n!r_n).$$
Note
$$frac 1{n+1}le n!r_n=frac 1{n+1}+frac 1{(n+1)(n+2)}+cdotslesum_{k=1}^inftyfrac1{(n+1)^k}=frac 1n.$$
By sandwich theorem, $n!r_nto 0$ and $ncdot n!r_nto 1$ as $ntoinfty$. Using $sinthetasimtheta$ as $thetato 0$, we get
$$lim_{ntoinfty}nsin(2pi n! e)=lim_{ntoinfty}nsin(2pi n! r_n)=2pi.$$
To prove $e$ irrational, assume $e=p/q$, then when $n>q$, $n!einBbb Z$ implies $sin(2pi n! e)equiv 0$, and
$$lim_{ntoinfty}nsin(2pi n! e)=lim_{ntoinfty}ncdot 0=0ne 2pi,$$
a contradiction.
answered Nov 18 at 13:03
Tianlalu
2,8911935
I love you :), you made it clear, thanks!
– Lamethyste
Nov 18 at 13:07
The key idea here is the bound for $r_n$. If $e$ were rational say $m/n$ then $n! e=I_n+n! r_n$ is an integer and this is a contradiction as $n! r_nin[1/(n+1),1/n] $.
– Paramanand Singh
Nov 18 at 14:07
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $I_n=sumlimits_{k=0}^ndfrac{n!}{k!}in Bbb Z$, then
$$nsin(2pi n! e)=nsin(2pi I_n+2pi n!r_n)=nsin(2pi n!r_n).$$
Note
$$frac 1{n+1}le n!r_n=frac 1{n+1}+frac 1{(n+1)(n+2)}+cdotslesum_{k=1}^inftyfrac1{(n+1)^k}=frac 1n.$$
By sandwich theorem, $n!r_nto 0$ and $ncdot n!r_nto 1$ as $ntoinfty$. Using $sinthetasimtheta$ as $thetato 0$, we get
$$lim_{ntoinfty}nsin(2pi n! e)=lim_{ntoinfty}nsin(2pi n! r_n)=2pi.$$
To prove $e$ irrational, assume $e=p/q$, then when $n>q$, $n!einBbb Z$ implies $sin(2pi n! e)equiv 0$, and
$$lim_{ntoinfty}nsin(2pi n! e)=lim_{ntoinfty}ncdot 0=0ne 2pi,$$
a contradiction.
answered Nov 18 at 13:03
Tianlalu
2,8911935
I love you :), you made it clear, thanks!
– Lamethyste
Nov 18 at 13:07
The key idea here is the bound for $r_n$. If $e$ were rational say $m/n$ then $n! e=I_n+n! r_n$ is an integer and this is a contradiction as $n! r_nin[1/(n+1),1/n] $.
– Paramanand Singh
Nov 18 at 14:07
add a comment |
up vote
1
down vote
accepted
Let $I_n=sumlimits_{k=0}^ndfrac{n!}{k!}in Bbb Z$, then
$$nsin(2pi n! e)=nsin(2pi I_n+2pi n!r_n)=nsin(2pi n!r_n).$$
Note
$$frac 1{n+1}le n!r_n=frac 1{n+1}+frac 1{(n+1)(n+2)}+cdotslesum_{k=1}^inftyfrac1{(n+1)^k}=frac 1n.$$
By sandwich theorem, $n!r_nto 0$ and $ncdot n!r_nto 1$ as $ntoinfty$. Using $sinthetasimtheta$ as $thetato 0$, we get
$$lim_{ntoinfty}nsin(2pi n! e)=lim_{ntoinfty}nsin(2pi n! r_n)=2pi.$$
To prove $e$ irrational, assume $e=p/q$, then when $n>q$, $n!einBbb Z$ implies $sin(2pi n! e)equiv 0$, and
$$lim_{ntoinfty}nsin(2pi n! e)=lim_{ntoinfty}ncdot 0=0ne 2pi,$$
a contradiction.
answered Nov 18 at 13:03
Tianlalu
2,8911935
I love you :), you made it clear, thanks!
– Lamethyste
Nov 18 at 13:07
The key idea here is the bound for $r_n$. If $e$ were rational say $m/n$ then $n! e=I_n+n! r_n$ is an integer and this is a contradiction as $n! r_nin[1/(n+1),1/n] $.
– Paramanand Singh
Nov 18 at 14:07
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $I_n=sumlimits_{k=0}^ndfrac{n!}{k!}in Bbb Z$, then
$$nsin(2pi n! e)=nsin(2pi I_n+2pi n!r_n)=nsin(2pi n!r_n).$$
Note
$$frac 1{n+1}le n!r_n=frac 1{n+1}+frac 1{(n+1)(n+2)}+cdotslesum_{k=1}^inftyfrac1{(n+1)^k}=frac 1n.$$
By sandwich theorem, $n!r_nto 0$ and $ncdot n!r_nto 1$ as $ntoinfty$. Using $sinthetasimtheta$ as $thetato 0$, we get
$$lim_{ntoinfty}nsin(2pi n! e)=lim_{ntoinfty}nsin(2pi n! r_n)=2pi.$$
To prove $e$ irrational, assume $e=p/q$, then when $n>q$, $n!einBbb Z$ implies $sin(2pi n! e)equiv 0$, and
$$lim_{ntoinfty}nsin(2pi n! e)=lim_{ntoinfty}ncdot 0=0ne 2pi,$$
a contradiction.
answered Nov 18 at 13:03
Tianlalu
2,8911935
Let $I_n=sumlimits_{k=0}^ndfrac{n!}{k!}in Bbb Z$, then
$$nsin(2pi n! e)=nsin(2pi I_n+2pi n!r_n)=nsin(2pi n!r_n).$$
Note
$$frac 1{n+1}le n!r_n=frac 1{n+1}+frac 1{(n+1)(n+2)}+cdotslesum_{k=1}^inftyfrac1{(n+1)^k}=frac 1n.$$
By sandwich theorem, $n!r_nto 0$ and $ncdot n!r_nto 1$ as $ntoinfty$. Using $sinthetasimtheta$ as $thetato 0$, we get
$$lim_{ntoinfty}nsin(2pi n! e)=lim_{ntoinfty}nsin(2pi n! r_n)=2pi.$$
To prove $e$ irrational, assume $e=p/q$, then when $n>q$, $n!einBbb Z$ implies $sin(2pi n! e)equiv 0$, and
$$lim_{ntoinfty}nsin(2pi n! e)=lim_{ntoinfty}ncdot 0=0ne 2pi,$$
a contradiction.
answered Nov 18 at 13:03
Tianlalu
2,8911935
edited Nov 18 at 13:12
answered Nov 18 at 13:03
Tianlalu
2,8911935
answered Nov 18 at 13:03
Tianlalu
2,8911935
answered Nov 18 at 13:03
Tianlalu
2,8911935
2,8911935
I love you :), you made it clear, thanks!
– Lamethyste
Nov 18 at 13:07
The key idea here is the bound for $r_n$. If $e$ were rational say $m/n$ then $n! e=I_n+n! r_n$ is an integer and this is a contradiction as $n! r_nin[1/(n+1),1/n] $.
– Paramanand Singh
Nov 18 at 14:07
add a comment |
I love you :), you made it clear, thanks!
– Lamethyste
Nov 18 at 13:07
The key idea here is the bound for $r_n$. If $e$ were rational say $m/n$ then $n! e=I_n+n! r_n$ is an integer and this is a contradiction as $n! r_nin[1/(n+1),1/n] $.
– Paramanand Singh
Nov 18 at 14:07
I love you :), you made it clear, thanks!
– Lamethyste
Nov 18 at 13:07
I love you :), you made it clear, thanks!
– Lamethyste
Nov 18 at 13:07
The key idea here is the bound for $r_n$. If $e$ were rational say $m/n$ then $n! e=I_n+n! r_n$ is an integer and this is a contradiction as $n! r_nin[1/(n+1),1/n] $.
– Paramanand Singh
Nov 18 at 14:07
The key idea here is the bound for $r_n$. If $e$ were rational say $m/n$ then $n! e=I_n+n! r_n$ is an integer and this is a contradiction as $n! r_nin[1/(n+1),1/n] $.
– Paramanand Singh
Nov 18 at 14:07
add a comment |
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Well, ok. The first limit seems manageable. What are your thoughts?
– Yuriy S
Nov 18 at 12:17
I tried by composition because $limlimits_{ntoinfty} 2πn!r_n = 0$ and I showed $$limlimits_{ntoinfty} nsin(2πn!r_n)= 2π$$ but i don't think that's correct
– Lamethyste
Nov 18 at 12:26
Seems fine to me. But not sure about the other limit
– Yuriy S
Nov 18 at 12:48
Shouldn't the sum in the definition of $r_n$ start at $k=0$ instead of $k=1$?
– Barry Cipra
Nov 18 at 12:50
You are right, my bad!
– Lamethyste
Nov 18 at 12:55