Inequality deduced from relatively prime numbers.












3












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















3












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














3












3








3


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 6:50









Alex Ravsky

42.3k32383




42.3k32383










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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes this is exactly what I was looking for - it seems rather obvious now!
    $endgroup$
    – An Invisible Carrot
    Dec 26 '18 at 8:57



















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3048194%2finequality-deduced-from-relatively-prime-numbers%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes this is exactly what I was looking for - it seems rather obvious now!
    $endgroup$
    – An Invisible Carrot
    Dec 26 '18 at 8:57
















1












$begingroup$

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!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes this is exactly what I was looking for - it seems rather obvious now!
    $endgroup$
    – An Invisible Carrot
    Dec 26 '18 at 8:57














1












1








1





$begingroup$

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!






share|cite|improve this answer









$endgroup$



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!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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











0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 22 '18 at 6:49









Alex RavskyAlex Ravsky

42.3k32383




42.3k32383












  • $begingroup$
    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










  • $begingroup$
    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 $b_{n+1}>b_n$)
$endgroup$
– Hugh Entwistle
Dec 24 '18 at 23:57


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3048194%2finequality-deduced-from-relatively-prime-numbers%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Wiesbaden

Image
Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
Read more

Marschland

Image
Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
Read more

Dieringhausen

Image
Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
Read more