Inequality deduced from relatively prime numbers.
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 at 6:50
Alex Ravsky
38.5k32079
asked Dec 21 at 5:05
An Invisible Carrot
998
|
show 1 more comment
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 at 6:50
Alex Ravsky
38.5k32079
asked Dec 21 at 5:05
An Invisible Carrot
998
That series looks like it converges to an irrational.
– Lord Shark the Unknown
Dec 21 at 5:08
1
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!
– An Invisible Carrot
Dec 21 at 5:11
Can't you just adapt the usual irrationality proof for $e$?
– Lord Shark the Unknown
Dec 21 at 5:13
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.
– An Invisible Carrot
Dec 21 at 5:17
@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?
– Suraj
Dec 21 at 6:19
|
show 1 more comment
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 at 6:50
Alex Ravsky
38.5k32079
asked Dec 21 at 5:05
An Invisible Carrot
998
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 at 6:50
Alex Ravsky
38.5k32079
asked Dec 21 at 5:05
An Invisible Carrot
998
edited Dec 22 at 6:50
Alex Ravsky
38.5k32079
asked Dec 21 at 5:05
An Invisible Carrot
998
edited Dec 22 at 6:50
Alex Ravsky
38.5k32079
edited Dec 22 at 6:50
Alex Ravsky
38.5k32079
edited Dec 22 at 6:50
Alex Ravsky
38.5k32079
38.5k32079
asked Dec 21 at 5:05
An Invisible Carrot
998
asked Dec 21 at 5:05
An Invisible Carrot
998
asked Dec 21 at 5:05
An Invisible Carrot
998
998
That series looks like it converges to an irrational.
– Lord Shark the Unknown
Dec 21 at 5:08
1
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!
– An Invisible Carrot
Dec 21 at 5:11
Can't you just adapt the usual irrationality proof for $e$?
– Lord Shark the Unknown
Dec 21 at 5:13
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.
– An Invisible Carrot
Dec 21 at 5:17
@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?
– Suraj
Dec 21 at 6:19
|
show 1 more comment
That series looks like it converges to an irrational.
– Lord Shark the Unknown
Dec 21 at 5:08
1
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!
– An Invisible Carrot
Dec 21 at 5:11
Can't you just adapt the usual irrationality proof for $e$?
– Lord Shark the Unknown
Dec 21 at 5:13
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.
– An Invisible Carrot
Dec 21 at 5:17
@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?
– Suraj
Dec 21 at 6:19
That series looks like it converges to an irrational.
– Lord Shark the Unknown
Dec 21 at 5:08
That series looks like it converges to an irrational.
– Lord Shark the Unknown
Dec 21 at 5:08
1
1
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!
– An Invisible Carrot
Dec 21 at 5:11
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!
– An Invisible Carrot
Dec 21 at 5:11
Can't you just adapt the usual irrationality proof for $e$?
– Lord Shark the Unknown
Dec 21 at 5:13
Can't you just adapt the usual irrationality proof for $e$?
– Lord Shark the Unknown
Dec 21 at 5:13
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.
– An Invisible Carrot
Dec 21 at 5:17
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.
– An Invisible Carrot
Dec 21 at 5:17
@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?
– Suraj
Dec 21 at 6:19
@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?
– Suraj
Dec 21 at 6:19
|
show 1 more comment
2 Answers
2
active
oldest
votes
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 at 6:49
Alex Ravsky
38.5k32079
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
– An Invisible Carrot
23 hours ago
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
– Hugh Entwistle
23 hours ago
add a comment |
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 23 hours ago
Hugh Entwistle
620216
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 at 6:49
Alex Ravsky
38.5k32079
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
– An Invisible Carrot
23 hours ago
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
– Hugh Entwistle
23 hours ago
add a comment |
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 at 6:49
Alex Ravsky
38.5k32079
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
– An Invisible Carrot
23 hours ago
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
– Hugh Entwistle
23 hours ago
add a comment |
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 at 6:49
Alex Ravsky
38.5k32079
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 at 6:49
Alex Ravsky
38.5k32079
answered Dec 22 at 6:49
Alex Ravsky
38.5k32079
answered Dec 22 at 6:49
Alex Ravsky
38.5k32079
answered Dec 22 at 6:49
Alex Ravsky
38.5k32079
38.5k32079
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
– An Invisible Carrot
23 hours ago
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
– Hugh Entwistle
23 hours ago
add a comment |
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
– An Invisible Carrot
23 hours ago
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
– Hugh Entwistle
23 hours ago
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
– An Invisible Carrot
23 hours ago
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
– An Invisible Carrot
23 hours ago
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
– Hugh Entwistle
23 hours ago
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
– Hugh Entwistle
23 hours ago
add a comment |
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 23 hours ago
Hugh Entwistle
620216
add a comment |
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 23 hours ago
Hugh Entwistle
620216
add a comment |
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 23 hours ago
Hugh Entwistle
620216
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 23 hours ago
Hugh Entwistle
620216
answered 23 hours ago
Hugh Entwistle
620216
answered 23 hours ago
Hugh Entwistle
620216
answered 23 hours ago
Hugh Entwistle
620216
620216
add a comment |
add a comment |
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That series looks like it converges to an irrational.
– Lord Shark the Unknown
Dec 21 at 5:08
1
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!
– An Invisible Carrot
Dec 21 at 5:11
Can't you just adapt the usual irrationality proof for $e$?
– Lord Shark the Unknown
Dec 21 at 5:13
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.
– An Invisible Carrot
Dec 21 at 5:17
@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?
– Suraj
Dec 21 at 6:19