Inequality deduced from relatively prime numbers.
$begingroup$
If $a_n text{ and }b_n$ are relatively prime for all $n$ and
$$frac{a_n}{b_n}=frac{1}{n}+frac{1}{n(n+1)}+frac{1}{n(n+1)(n+2)}+cdots$$
Deduce that
$$b_ngeq b_{n+1}$$
CURRENT THOUGHTS
I can show that
$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$
and making $b_n$ the subject
$$b_n=(frac{na_n}{a_{n+1}+b_{n+1}})b_{n+1}$$
so it would suffice to show that
$$na_ngeq a_{n+1}+b_{n+1}$$
which would appear to be true in general as $n$ gets large. But other than this, I am unsure how to proceed?
inequality prime-numbers irrational-numbers
edited Dec 22 '18 at 6:50
Alex Ravsky
42.4k32383
asked Dec 21 '18 at 5:05
An Invisible CarrotAn Invisible Carrot
1018
$endgroup$
|
show 1 more comment
$begingroup$
If $a_n text{ and }b_n$ are relatively prime for all $n$ and
$$frac{a_n}{b_n}=frac{1}{n}+frac{1}{n(n+1)}+frac{1}{n(n+1)(n+2)}+cdots$$
Deduce that
$$b_ngeq b_{n+1}$$
CURRENT THOUGHTS
I can show that
$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$
and making $b_n$ the subject
$$b_n=(frac{na_n}{a_{n+1}+b_{n+1}})b_{n+1}$$
so it would suffice to show that
$$na_ngeq a_{n+1}+b_{n+1}$$
which would appear to be true in general as $n$ gets large. But other than this, I am unsure how to proceed?
inequality prime-numbers irrational-numbers
edited Dec 22 '18 at 6:50
Alex Ravsky
42.4k32383
asked Dec 21 '18 at 5:05
An Invisible CarrotAn Invisible Carrot
1018
$endgroup$
$begingroup$
That series looks like it converges to an irrational.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:08
1
$begingroup$
Indeed it does, I am trying to write an elementary proof of this for my students -- if I can show this inequality then it will follow that the series is irrational. But I was hoping to not use the fact that it is irrational to prove the inequality!
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:11
$begingroup$
Can't you just adapt the usual irrationality proof for $e$?
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:13
$begingroup$
I dont know what you mean by 'usual' proof, but I am currently establishing $x_n=frac{a_n}{b_n}$ where $e$ being rational would imply that all $x_n$ are rational, and then showing that $x_n$ is not rational by contradiction. The inequality $b_n geq b_{n+1}$ is all I need.
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:17
$begingroup$
@AnInvisibleCarrot I wonder how to get $a_n$and $b_n$ when it doesn't really exists because your RHS seem to converge to an irrational?
$endgroup$
– user408906
Dec 21 '18 at 6:19
|
show 1 more comment
$begingroup$
If $a_n text{ and }b_n$ are relatively prime for all $n$ and
$$frac{a_n}{b_n}=frac{1}{n}+frac{1}{n(n+1)}+frac{1}{n(n+1)(n+2)}+cdots$$
Deduce that
$$b_ngeq b_{n+1}$$
CURRENT THOUGHTS
I can show that
$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$
and making $b_n$ the subject
$$b_n=(frac{na_n}{a_{n+1}+b_{n+1}})b_{n+1}$$
so it would suffice to show that
$$na_ngeq a_{n+1}+b_{n+1}$$
which would appear to be true in general as $n$ gets large. But other than this, I am unsure how to proceed?
inequality prime-numbers irrational-numbers
edited Dec 22 '18 at 6:50
Alex Ravsky
42.4k32383
asked Dec 21 '18 at 5:05
An Invisible CarrotAn Invisible Carrot
1018
$endgroup$
If $a_n text{ and }b_n$ are relatively prime for all $n$ and
$$frac{a_n}{b_n}=frac{1}{n}+frac{1}{n(n+1)}+frac{1}{n(n+1)(n+2)}+cdots$$
Deduce that
$$b_ngeq b_{n+1}$$
CURRENT THOUGHTS
I can show that
$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$
and making $b_n$ the subject
$$b_n=(frac{na_n}{a_{n+1}+b_{n+1}})b_{n+1}$$
so it would suffice to show that
$$na_ngeq a_{n+1}+b_{n+1}$$
which would appear to be true in general as $n$ gets large. But other than this, I am unsure how to proceed?
inequality prime-numbers irrational-numbers
inequality prime-numbers irrational-numbers
edited Dec 22 '18 at 6:50
Alex Ravsky
42.4k32383
asked Dec 21 '18 at 5:05
An Invisible CarrotAn Invisible Carrot
1018
edited Dec 22 '18 at 6:50
Alex Ravsky
42.4k32383
asked Dec 21 '18 at 5:05
An Invisible CarrotAn Invisible Carrot
1018
edited Dec 22 '18 at 6:50
Alex Ravsky
42.4k32383
edited Dec 22 '18 at 6:50
Alex Ravsky
42.4k32383
edited Dec 22 '18 at 6:50
Alex Ravsky
42.4k32383
42.4k32383
asked Dec 21 '18 at 5:05
An Invisible CarrotAn Invisible Carrot
1018
asked Dec 21 '18 at 5:05
An Invisible CarrotAn Invisible Carrot
1018
asked Dec 21 '18 at 5:05
An Invisible CarrotAn Invisible Carrot
1018
1018
$begingroup$
That series looks like it converges to an irrational.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:08
1
$begingroup$
Indeed it does, I am trying to write an elementary proof of this for my students -- if I can show this inequality then it will follow that the series is irrational. But I was hoping to not use the fact that it is irrational to prove the inequality!
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:11
$begingroup$
Can't you just adapt the usual irrationality proof for $e$?
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:13
$begingroup$
I dont know what you mean by 'usual' proof, but I am currently establishing $x_n=frac{a_n}{b_n}$ where $e$ being rational would imply that all $x_n$ are rational, and then showing that $x_n$ is not rational by contradiction. The inequality $b_n geq b_{n+1}$ is all I need.
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:17
$begingroup$
@AnInvisibleCarrot I wonder how to get $a_n$and $b_n$ when it doesn't really exists because your RHS seem to converge to an irrational?
$endgroup$
– user408906
Dec 21 '18 at 6:19
|
show 1 more comment
$begingroup$
That series looks like it converges to an irrational.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:08
1
$begingroup$
Indeed it does, I am trying to write an elementary proof of this for my students -- if I can show this inequality then it will follow that the series is irrational. But I was hoping to not use the fact that it is irrational to prove the inequality!
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:11
$begingroup$
Can't you just adapt the usual irrationality proof for $e$?
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:13
$begingroup$
I dont know what you mean by 'usual' proof, but I am currently establishing $x_n=frac{a_n}{b_n}$ where $e$ being rational would imply that all $x_n$ are rational, and then showing that $x_n$ is not rational by contradiction. The inequality $b_n geq b_{n+1}$ is all I need.
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:17
$begingroup$
@AnInvisibleCarrot I wonder how to get $a_n$and $b_n$ when it doesn't really exists because your RHS seem to converge to an irrational?
$endgroup$
– user408906
Dec 21 '18 at 6:19
$begingroup$
That series looks like it converges to an irrational.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:08
$begingroup$
That series looks like it converges to an irrational.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:08
1
1
$begingroup$
Indeed it does, I am trying to write an elementary proof of this for my students -- if I can show this inequality then it will follow that the series is irrational. But I was hoping to not use the fact that it is irrational to prove the inequality!
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:11
$begingroup$
Indeed it does, I am trying to write an elementary proof of this for my students -- if I can show this inequality then it will follow that the series is irrational. But I was hoping to not use the fact that it is irrational to prove the inequality!
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:11
$begingroup$
Can't you just adapt the usual irrationality proof for $e$?
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:13
$begingroup$
Can't you just adapt the usual irrationality proof for $e$?
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:13
$begingroup$
I dont know what you mean by 'usual' proof, but I am currently establishing $x_n=frac{a_n}{b_n}$ where $e$ being rational would imply that all $x_n$ are rational, and then showing that $x_n$ is not rational by contradiction. The inequality $b_n geq b_{n+1}$ is all I need.
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:17
$begingroup$
I dont know what you mean by 'usual' proof, but I am currently establishing $x_n=frac{a_n}{b_n}$ where $e$ being rational would imply that all $x_n$ are rational, and then showing that $x_n$ is not rational by contradiction. The inequality $b_n geq b_{n+1}$ is all I need.
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:17
$begingroup$
@AnInvisibleCarrot I wonder how to get $a_n$and $b_n$ when it doesn't really exists because your RHS seem to converge to an irrational?
$endgroup$
– user408906
Dec 21 '18 at 6:19
$begingroup$
@AnInvisibleCarrot I wonder how to get $a_n$and $b_n$ when it doesn't really exists because your RHS seem to converge to an irrational?
$endgroup$
– user408906
Dec 21 '18 at 6:19
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
I believe this statement naturally follows from the fact that $a_n$ and $b_n$ are defined to be coprime for all $n$.
i.e.
If
$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$
Then we must have that $b_n geq b_{n+1}$ otherwise $a_{n+1}$ and $b_{n+1}$ wouldn't be coprime.
Hope this helps!
If I am not mistaken, the next steps in this proof will be to conclude that
$$frac{a_n}{b_n}>frac{a_{n+1}}{b_{n+1}}Rightarrow a_{n+1}<a_n$$
And we obtain a contradiction by
$$a_1>a_2>a_3>cdots >0$$
Since it is not possible to have a infinite decreasing sequence of positive integers?
Very nice!
answered Dec 25 '18 at 0:21
Hugh EntwistleHugh Entwistle
841217
$endgroup$
$begingroup$
Yes this is exactly what I was looking for - it seems rather obvious now!
$endgroup$
– An Invisible Carrot
Dec 26 '18 at 8:57
add a comment |
$begingroup$
There is a simple straightforward proof that for each $n$ $$S(n)=sum_{i=n}^inftyprod_{j=n}^i frac 1j$$ is irrational. Indeed, suppose to the contrary that $S(n)=frac {a_n}{b_n}$ where $a_n$ and $b_n$ are integers. Put $m=n+b_n$. Among $b_n$ consecutive numbers $n, n+1,dots, m-1$ exists one divisible by $b_n$. Then $$T(n)=n(n+1)cdotcdotscdot (m-1)S(n)$$ is an integer. On the other hand, $T(n)$ is a sum of an integer and $S(m)$. But
$$0<S(m)<sum_{i=m}^infty m^{m-i-1}=frac 1mcdotfrac{1}{1-frac 1{m}}=frac1{m-1}<1,$$
a contradiction.
answered Dec 22 '18 at 6:49
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
$begingroup$
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
$endgroup$
– An Invisible Carrot
Dec 24 '18 at 23:53
$begingroup$
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
$endgroup$
– Hugh Entwistle
Dec 24 '18 at 23:57
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I believe this statement naturally follows from the fact that $a_n$ and $b_n$ are defined to be coprime for all $n$.
i.e.
If
$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$
Then we must have that $b_n geq b_{n+1}$ otherwise $a_{n+1}$ and $b_{n+1}$ wouldn't be coprime.
Hope this helps!
If I am not mistaken, the next steps in this proof will be to conclude that
$$frac{a_n}{b_n}>frac{a_{n+1}}{b_{n+1}}Rightarrow a_{n+1}<a_n$$
And we obtain a contradiction by
$$a_1>a_2>a_3>cdots >0$$
Since it is not possible to have a infinite decreasing sequence of positive integers?
Very nice!
answered Dec 25 '18 at 0:21
Hugh EntwistleHugh Entwistle
841217
$endgroup$
$begingroup$
Yes this is exactly what I was looking for - it seems rather obvious now!
$endgroup$
– An Invisible Carrot
Dec 26 '18 at 8:57
add a comment |
$begingroup$
I believe this statement naturally follows from the fact that $a_n$ and $b_n$ are defined to be coprime for all $n$.
i.e.
If
$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$
Then we must have that $b_n geq b_{n+1}$ otherwise $a_{n+1}$ and $b_{n+1}$ wouldn't be coprime.
Hope this helps!
If I am not mistaken, the next steps in this proof will be to conclude that
$$frac{a_n}{b_n}>frac{a_{n+1}}{b_{n+1}}Rightarrow a_{n+1}<a_n$$
And we obtain a contradiction by
$$a_1>a_2>a_3>cdots >0$$
Since it is not possible to have a infinite decreasing sequence of positive integers?
Very nice!
answered Dec 25 '18 at 0:21
Hugh EntwistleHugh Entwistle
841217
$endgroup$
$begingroup$
Yes this is exactly what I was looking for - it seems rather obvious now!
$endgroup$
– An Invisible Carrot
Dec 26 '18 at 8:57
add a comment |
$begingroup$
I believe this statement naturally follows from the fact that $a_n$ and $b_n$ are defined to be coprime for all $n$.
i.e.
If
$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$
Then we must have that $b_n geq b_{n+1}$ otherwise $a_{n+1}$ and $b_{n+1}$ wouldn't be coprime.
Hope this helps!
If I am not mistaken, the next steps in this proof will be to conclude that
$$frac{a_n}{b_n}>frac{a_{n+1}}{b_{n+1}}Rightarrow a_{n+1}<a_n$$
And we obtain a contradiction by
$$a_1>a_2>a_3>cdots >0$$
Since it is not possible to have a infinite decreasing sequence of positive integers?
Very nice!
answered Dec 25 '18 at 0:21
Hugh EntwistleHugh Entwistle
841217
$endgroup$
I believe this statement naturally follows from the fact that $a_n$ and $b_n$ are defined to be coprime for all $n$.
i.e.
If
$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$
Then we must have that $b_n geq b_{n+1}$ otherwise $a_{n+1}$ and $b_{n+1}$ wouldn't be coprime.
Hope this helps!
If I am not mistaken, the next steps in this proof will be to conclude that
$$frac{a_n}{b_n}>frac{a_{n+1}}{b_{n+1}}Rightarrow a_{n+1}<a_n$$
And we obtain a contradiction by
$$a_1>a_2>a_3>cdots >0$$
Since it is not possible to have a infinite decreasing sequence of positive integers?
Very nice!
answered Dec 25 '18 at 0:21
Hugh EntwistleHugh Entwistle
841217
answered Dec 25 '18 at 0:21
Hugh EntwistleHugh Entwistle
841217
answered Dec 25 '18 at 0:21
Hugh EntwistleHugh Entwistle
841217
answered Dec 25 '18 at 0:21
Hugh EntwistleHugh Entwistle
841217
841217
$begingroup$
Yes this is exactly what I was looking for - it seems rather obvious now!
$endgroup$
– An Invisible Carrot
Dec 26 '18 at 8:57
add a comment |
$begingroup$
Yes this is exactly what I was looking for - it seems rather obvious now!
$endgroup$
– An Invisible Carrot
Dec 26 '18 at 8:57
$begingroup$
Yes this is exactly what I was looking for - it seems rather obvious now!
$endgroup$
– An Invisible Carrot
Dec 26 '18 at 8:57
$begingroup$
Yes this is exactly what I was looking for - it seems rather obvious now!
$endgroup$
– An Invisible Carrot
Dec 26 '18 at 8:57
add a comment |
$begingroup$
There is a simple straightforward proof that for each $n$ $$S(n)=sum_{i=n}^inftyprod_{j=n}^i frac 1j$$ is irrational. Indeed, suppose to the contrary that $S(n)=frac {a_n}{b_n}$ where $a_n$ and $b_n$ are integers. Put $m=n+b_n$. Among $b_n$ consecutive numbers $n, n+1,dots, m-1$ exists one divisible by $b_n$. Then $$T(n)=n(n+1)cdotcdotscdot (m-1)S(n)$$ is an integer. On the other hand, $T(n)$ is a sum of an integer and $S(m)$. But
$$0<S(m)<sum_{i=m}^infty m^{m-i-1}=frac 1mcdotfrac{1}{1-frac 1{m}}=frac1{m-1}<1,$$
a contradiction.
answered Dec 22 '18 at 6:49
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
$begingroup$
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
$endgroup$
– An Invisible Carrot
Dec 24 '18 at 23:53
$begingroup$
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
$endgroup$
– Hugh Entwistle
Dec 24 '18 at 23:57
add a comment |
$begingroup$
There is a simple straightforward proof that for each $n$ $$S(n)=sum_{i=n}^inftyprod_{j=n}^i frac 1j$$ is irrational. Indeed, suppose to the contrary that $S(n)=frac {a_n}{b_n}$ where $a_n$ and $b_n$ are integers. Put $m=n+b_n$. Among $b_n$ consecutive numbers $n, n+1,dots, m-1$ exists one divisible by $b_n$. Then $$T(n)=n(n+1)cdotcdotscdot (m-1)S(n)$$ is an integer. On the other hand, $T(n)$ is a sum of an integer and $S(m)$. But
$$0<S(m)<sum_{i=m}^infty m^{m-i-1}=frac 1mcdotfrac{1}{1-frac 1{m}}=frac1{m-1}<1,$$
a contradiction.
answered Dec 22 '18 at 6:49
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
$begingroup$
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
$endgroup$
– An Invisible Carrot
Dec 24 '18 at 23:53
$begingroup$
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
$endgroup$
– Hugh Entwistle
Dec 24 '18 at 23:57
add a comment |
$begingroup$
There is a simple straightforward proof that for each $n$ $$S(n)=sum_{i=n}^inftyprod_{j=n}^i frac 1j$$ is irrational. Indeed, suppose to the contrary that $S(n)=frac {a_n}{b_n}$ where $a_n$ and $b_n$ are integers. Put $m=n+b_n$. Among $b_n$ consecutive numbers $n, n+1,dots, m-1$ exists one divisible by $b_n$. Then $$T(n)=n(n+1)cdotcdotscdot (m-1)S(n)$$ is an integer. On the other hand, $T(n)$ is a sum of an integer and $S(m)$. But
$$0<S(m)<sum_{i=m}^infty m^{m-i-1}=frac 1mcdotfrac{1}{1-frac 1{m}}=frac1{m-1}<1,$$
a contradiction.
answered Dec 22 '18 at 6:49
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
There is a simple straightforward proof that for each $n$ $$S(n)=sum_{i=n}^inftyprod_{j=n}^i frac 1j$$ is irrational. Indeed, suppose to the contrary that $S(n)=frac {a_n}{b_n}$ where $a_n$ and $b_n$ are integers. Put $m=n+b_n$. Among $b_n$ consecutive numbers $n, n+1,dots, m-1$ exists one divisible by $b_n$. Then $$T(n)=n(n+1)cdotcdotscdot (m-1)S(n)$$ is an integer. On the other hand, $T(n)$ is a sum of an integer and $S(m)$. But
$$0<S(m)<sum_{i=m}^infty m^{m-i-1}=frac 1mcdotfrac{1}{1-frac 1{m}}=frac1{m-1}<1,$$
a contradiction.
answered Dec 22 '18 at 6:49
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 22 '18 at 6:49
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 22 '18 at 6:49
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 22 '18 at 6:49
Alex RavskyAlex Ravsky
42.4k32383
42.4k32383
$begingroup$
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
$endgroup$
– An Invisible Carrot
Dec 24 '18 at 23:53
$begingroup$
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
$endgroup$
– Hugh Entwistle
Dec 24 '18 at 23:57
add a comment |
$begingroup$
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
$endgroup$
– An Invisible Carrot
Dec 24 '18 at 23:53
$begingroup$
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
$endgroup$
– Hugh Entwistle
Dec 24 '18 at 23:57
$begingroup$
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
$endgroup$
– An Invisible Carrot
Dec 24 '18 at 23:53
$begingroup$
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
$endgroup$
– An Invisible Carrot
Dec 24 '18 at 23:53
$begingroup$
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
$endgroup$
– Hugh Entwistle
Dec 24 '18 at 23:57
$begingroup$
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
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– Hugh Entwistle
Dec 24 '18 at 23:57
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$begingroup$
That series looks like it converges to an irrational.
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– Lord Shark the Unknown
Dec 21 '18 at 5:08
1
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Indeed it does, I am trying to write an elementary proof of this for my students -- if I can show this inequality then it will follow that the series is irrational. But I was hoping to not use the fact that it is irrational to prove the inequality!
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– An Invisible Carrot
Dec 21 '18 at 5:11
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Can't you just adapt the usual irrationality proof for $e$?
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– Lord Shark the Unknown
Dec 21 '18 at 5:13
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I dont know what you mean by 'usual' proof, but I am currently establishing $x_n=frac{a_n}{b_n}$ where $e$ being rational would imply that all $x_n$ are rational, and then showing that $x_n$ is not rational by contradiction. The inequality $b_n geq b_{n+1}$ is all I need.
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– An Invisible Carrot
Dec 21 '18 at 5:17
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@AnInvisibleCarrot I wonder how to get $a_n$and $b_n$ when it doesn't really exists because your RHS seem to converge to an irrational?
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– user408906
Dec 21 '18 at 6:19