Mixing
Prove that the supremum of this set is $frac{1}{4}$: $S={frac{n-1}{4n+1}: n in mathbb{N}}$ without using...
$begingroup$
I’m trying to prove that the supremum of this set is $frac{1}{4}$: $S={frac{n-1}{4n+1}: n in mathbb{N}}$, without using limits.
I can easily show that $frac{1}{4}$ is an upper bound, but I have trouble showing that it is the least upper bound.
Strategy: 1) show it’s an upper bound by contradiction. 2) then, show that if there existed a lower bound $k$ that was strictly less than $frac{1}{4}$, then it wouldn’t be an upper bound because we could find an $n$ such that $frac{n-1}{4n+1}$ was greater than that upper bound. Do this by setting $frac{n-1}{4n+1}=1/4$, and then solving for $n$, as a function of $k$ and then incrementing it by one, and then showing that if we let $n$ = the function of $k$ incremented by one, then it would be greater than the upper bound $k$.
Proof:
Assume for contradiction that $frac{1}{4}$ is not an upper bound, then there exists $n$ such that $frac{n-1}{4n+1} > frac{1}{4}$, then since $4n+1 > 0$, $-4>1$ which is a contradiction. So it is an upper bound.
Assume for contradiction that there exists $k$ such that $k < frac{1}{4}$ and $k$ is an upper bound for $S$. Then since it’s an upper bound, $frac{n-1}{4n+1} le k$, which means $n le frac{k+1}{1-4 k }$, since $1-4k ge 0$.
Now I’m not sure where to go from here. I really want to use the Archimedean property, but since I need $n$ to be in $mathbb{N}$, I’m not sure if it is acceptable to use that. I’m also not sure what I can increment $n$ by as a function of $k$ in order to guarantee that it will still be in $mathbb{N}$. Can I get some hints maybe?
real-analysis analysis supremum-and-infimum
edited Jan 27 at 22:39
J. W. Tanner
3,8871320
asked Jan 27 at 21:49
jajajaja
8711
$endgroup$
add a comment |
$begingroup$
I’m trying to prove that the supremum of this set is $frac{1}{4}$: $S={frac{n-1}{4n+1}: n in mathbb{N}}$, without using limits.
I can easily show that $frac{1}{4}$ is an upper bound, but I have trouble showing that it is the least upper bound.
Strategy: 1) show it’s an upper bound by contradiction. 2) then, show that if there existed a lower bound $k$ that was strictly less than $frac{1}{4}$, then it wouldn’t be an upper bound because we could find an $n$ such that $frac{n-1}{4n+1}$ was greater than that upper bound. Do this by setting $frac{n-1}{4n+1}=1/4$, and then solving for $n$, as a function of $k$ and then incrementing it by one, and then showing that if we let $n$ = the function of $k$ incremented by one, then it would be greater than the upper bound $k$.
Proof:
Assume for contradiction that $frac{1}{4}$ is not an upper bound, then there exists $n$ such that $frac{n-1}{4n+1} > frac{1}{4}$, then since $4n+1 > 0$, $-4>1$ which is a contradiction. So it is an upper bound.
Assume for contradiction that there exists $k$ such that $k < frac{1}{4}$ and $k$ is an upper bound for $S$. Then since it’s an upper bound, $frac{n-1}{4n+1} le k$, which means $n le frac{k+1}{1-4 k }$, since $1-4k ge 0$.
Now I’m not sure where to go from here. I really want to use the Archimedean property, but since I need $n$ to be in $mathbb{N}$, I’m not sure if it is acceptable to use that. I’m also not sure what I can increment $n$ by as a function of $k$ in order to guarantee that it will still be in $mathbb{N}$. Can I get some hints maybe?
real-analysis analysis supremum-and-infimum
edited Jan 27 at 22:39
J. W. Tanner
3,8871320
asked Jan 27 at 21:49
jajajaja
8711
$endgroup$
add a comment |
$begingroup$
I’m trying to prove that the supremum of this set is $frac{1}{4}$: $S={frac{n-1}{4n+1}: n in mathbb{N}}$, without using limits.
I can easily show that $frac{1}{4}$ is an upper bound, but I have trouble showing that it is the least upper bound.
Strategy: 1) show it’s an upper bound by contradiction. 2) then, show that if there existed a lower bound $k$ that was strictly less than $frac{1}{4}$, then it wouldn’t be an upper bound because we could find an $n$ such that $frac{n-1}{4n+1}$ was greater than that upper bound. Do this by setting $frac{n-1}{4n+1}=1/4$, and then solving for $n$, as a function of $k$ and then incrementing it by one, and then showing that if we let $n$ = the function of $k$ incremented by one, then it would be greater than the upper bound $k$.
Proof:
Assume for contradiction that $frac{1}{4}$ is not an upper bound, then there exists $n$ such that $frac{n-1}{4n+1} > frac{1}{4}$, then since $4n+1 > 0$, $-4>1$ which is a contradiction. So it is an upper bound.
Assume for contradiction that there exists $k$ such that $k < frac{1}{4}$ and $k$ is an upper bound for $S$. Then since it’s an upper bound, $frac{n-1}{4n+1} le k$, which means $n le frac{k+1}{1-4 k }$, since $1-4k ge 0$.
Now I’m not sure where to go from here. I really want to use the Archimedean property, but since I need $n$ to be in $mathbb{N}$, I’m not sure if it is acceptable to use that. I’m also not sure what I can increment $n$ by as a function of $k$ in order to guarantee that it will still be in $mathbb{N}$. Can I get some hints maybe?
real-analysis analysis supremum-and-infimum
edited Jan 27 at 22:39
J. W. Tanner
3,8871320
asked Jan 27 at 21:49
jajajaja
8711
$endgroup$
I’m trying to prove that the supremum of this set is $frac{1}{4}$: $S={frac{n-1}{4n+1}: n in mathbb{N}}$, without using limits.
I can easily show that $frac{1}{4}$ is an upper bound, but I have trouble showing that it is the least upper bound.
Strategy: 1) show it’s an upper bound by contradiction. 2) then, show that if there existed a lower bound $k$ that was strictly less than $frac{1}{4}$, then it wouldn’t be an upper bound because we could find an $n$ such that $frac{n-1}{4n+1}$ was greater than that upper bound. Do this by setting $frac{n-1}{4n+1}=1/4$, and then solving for $n$, as a function of $k$ and then incrementing it by one, and then showing that if we let $n$ = the function of $k$ incremented by one, then it would be greater than the upper bound $k$.
Proof:
Assume for contradiction that $frac{1}{4}$ is not an upper bound, then there exists $n$ such that $frac{n-1}{4n+1} > frac{1}{4}$, then since $4n+1 > 0$, $-4>1$ which is a contradiction. So it is an upper bound.
Assume for contradiction that there exists $k$ such that $k < frac{1}{4}$ and $k$ is an upper bound for $S$. Then since it’s an upper bound, $frac{n-1}{4n+1} le k$, which means $n le frac{k+1}{1-4 k }$, since $1-4k ge 0$.
Now I’m not sure where to go from here. I really want to use the Archimedean property, but since I need $n$ to be in $mathbb{N}$, I’m not sure if it is acceptable to use that. I’m also not sure what I can increment $n$ by as a function of $k$ in order to guarantee that it will still be in $mathbb{N}$. Can I get some hints maybe?
real-analysis analysis supremum-and-infimum
real-analysis analysis supremum-and-infimum
edited Jan 27 at 22:39
J. W. Tanner
3,8871320
asked Jan 27 at 21:49
jajajaja
8711
edited Jan 27 at 22:39
J. W. Tanner
3,8871320
asked Jan 27 at 21:49
jajajaja
8711
edited Jan 27 at 22:39
J. W. Tanner
3,8871320
edited Jan 27 at 22:39
J. W. Tanner
3,8871320
edited Jan 27 at 22:39
J. W. Tanner
3,8871320
3,8871320
asked Jan 27 at 21:49
jajajaja
8711
asked Jan 27 at 21:49
jajajaja
8711
asked Jan 27 at 21:49
jajajaja
8711
8711
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In (1), the argument by contradiction is unnecessarily complicated. Instead, one could argue more simply that the numerator of $frac{n-1}{4n+1}$ is $< n$, and its denominator is $> 4n$, so this arbitrary element of $S$ is $< frac{n}{4n} = frac{1}{4}$.
In (2), appealing directly to the Archimedean property as you suggest (but also eschewing an argument by contradiction for a second time): given any $k < frac{1}{4}$, there exists a positive integer $n$ such that $nleft(frac{1}{4} - kright) > 1$, i.e. such that $k < frac{1}{4} - frac{1}{n} = frac{n-4}{4n}$. Conveniently, this fraction happens to be less than the element $frac{n-1}{4n+1}$ of $S$ (indeed, it is less than $frac{n-3}{4n+1}$, by the result that $frac{a}{b} < frac{a+1}{b+1}$ if $a < b$ and $b > 0$), so $k$ is not an upper bound of $S$.
answered Jan 27 at 23:44
Calum GilhooleyCalum Gilhooley
5,119630
$endgroup$
1
$begingroup$
(I'm braced for a ticking-off from @fleablood, for indulging in unnecessary cheese-paring!) :)
$endgroup$
– Calum Gilhooley
Jan 27 at 23:47
add a comment |
$begingroup$
Note that$$frac14-frac{n-1}{4n+1}=frac5{4(4n+1)}$$and$$frac5{4(4n+1)}<varepsiloniff n>frac{5-4varepsilon}{16varepsilon}.$$
answered Jan 27 at 21:56
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Thanks. However, that doesn't show me a methodical way of reaching that conclusion.
$endgroup$
– jaja
Jan 28 at 0:00
add a comment |
$begingroup$
Well, if $s = frac {n-1}{4n+1} in S; nin mathbb N$ then $frac {n-1}{4n+1} = frac {n+frac 14}{4n+1} - frac {frac 54}{4n+1} = frac 14 - frac {5}{16n + 4}< frac 14$.
So $frac 14$ is an upperbound of $S$.
If $k < frac 14$ and if $epsilon = frac 14 -k>0$.
$frac 5{16n + 4} < epsilon iff$
$frac 5{epsilon} < 16n + 4 iff$
$n > frac {frac 5epsilon -4}{16}$.
Such an $n$ can always be found and if we let $n > frac {frac 5epsilon -4}{16}$ we will have:
$k = frac 14 - epsilon < frac 14 - frac 5{16n+4} =frac {n-1}{4n+1}in S$.
So $k$ is not an upper bound of $S$.
And that is the definition of least upper bound. So $frac 14 =sup S$.
answered Jan 27 at 22:11
fleabloodfleablood
73.4k22791
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090161%2fprove-that-the-supremum-of-this-set-is-frac14-s-fracn-14n1-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In (1), the argument by contradiction is unnecessarily complicated. Instead, one could argue more simply that the numerator of $frac{n-1}{4n+1}$ is $< n$, and its denominator is $> 4n$, so this arbitrary element of $S$ is $< frac{n}{4n} = frac{1}{4}$.
In (2), appealing directly to the Archimedean property as you suggest (but also eschewing an argument by contradiction for a second time): given any $k < frac{1}{4}$, there exists a positive integer $n$ such that $nleft(frac{1}{4} - kright) > 1$, i.e. such that $k < frac{1}{4} - frac{1}{n} = frac{n-4}{4n}$. Conveniently, this fraction happens to be less than the element $frac{n-1}{4n+1}$ of $S$ (indeed, it is less than $frac{n-3}{4n+1}$, by the result that $frac{a}{b} < frac{a+1}{b+1}$ if $a < b$ and $b > 0$), so $k$ is not an upper bound of $S$.
answered Jan 27 at 23:44
Calum GilhooleyCalum Gilhooley
5,119630
$endgroup$
1
$begingroup$
(I'm braced for a ticking-off from @fleablood, for indulging in unnecessary cheese-paring!) :)
$endgroup$
– Calum Gilhooley
Jan 27 at 23:47
add a comment |
$begingroup$
In (1), the argument by contradiction is unnecessarily complicated. Instead, one could argue more simply that the numerator of $frac{n-1}{4n+1}$ is $< n$, and its denominator is $> 4n$, so this arbitrary element of $S$ is $< frac{n}{4n} = frac{1}{4}$.
In (2), appealing directly to the Archimedean property as you suggest (but also eschewing an argument by contradiction for a second time): given any $k < frac{1}{4}$, there exists a positive integer $n$ such that $nleft(frac{1}{4} - kright) > 1$, i.e. such that $k < frac{1}{4} - frac{1}{n} = frac{n-4}{4n}$. Conveniently, this fraction happens to be less than the element $frac{n-1}{4n+1}$ of $S$ (indeed, it is less than $frac{n-3}{4n+1}$, by the result that $frac{a}{b} < frac{a+1}{b+1}$ if $a < b$ and $b > 0$), so $k$ is not an upper bound of $S$.
answered Jan 27 at 23:44
Calum GilhooleyCalum Gilhooley
5,119630
$endgroup$
1
$begingroup$
(I'm braced for a ticking-off from @fleablood, for indulging in unnecessary cheese-paring!) :)
$endgroup$
– Calum Gilhooley
Jan 27 at 23:47
add a comment |
$begingroup$
In (1), the argument by contradiction is unnecessarily complicated. Instead, one could argue more simply that the numerator of $frac{n-1}{4n+1}$ is $< n$, and its denominator is $> 4n$, so this arbitrary element of $S$ is $< frac{n}{4n} = frac{1}{4}$.
In (2), appealing directly to the Archimedean property as you suggest (but also eschewing an argument by contradiction for a second time): given any $k < frac{1}{4}$, there exists a positive integer $n$ such that $nleft(frac{1}{4} - kright) > 1$, i.e. such that $k < frac{1}{4} - frac{1}{n} = frac{n-4}{4n}$. Conveniently, this fraction happens to be less than the element $frac{n-1}{4n+1}$ of $S$ (indeed, it is less than $frac{n-3}{4n+1}$, by the result that $frac{a}{b} < frac{a+1}{b+1}$ if $a < b$ and $b > 0$), so $k$ is not an upper bound of $S$.
answered Jan 27 at 23:44
Calum GilhooleyCalum Gilhooley
5,119630
$endgroup$
In (1), the argument by contradiction is unnecessarily complicated. Instead, one could argue more simply that the numerator of $frac{n-1}{4n+1}$ is $< n$, and its denominator is $> 4n$, so this arbitrary element of $S$ is $< frac{n}{4n} = frac{1}{4}$.
In (2), appealing directly to the Archimedean property as you suggest (but also eschewing an argument by contradiction for a second time): given any $k < frac{1}{4}$, there exists a positive integer $n$ such that $nleft(frac{1}{4} - kright) > 1$, i.e. such that $k < frac{1}{4} - frac{1}{n} = frac{n-4}{4n}$. Conveniently, this fraction happens to be less than the element $frac{n-1}{4n+1}$ of $S$ (indeed, it is less than $frac{n-3}{4n+1}$, by the result that $frac{a}{b} < frac{a+1}{b+1}$ if $a < b$ and $b > 0$), so $k$ is not an upper bound of $S$.
answered Jan 27 at 23:44
Calum GilhooleyCalum Gilhooley
5,119630
answered Jan 27 at 23:44
Calum GilhooleyCalum Gilhooley
5,119630
answered Jan 27 at 23:44
Calum GilhooleyCalum Gilhooley
5,119630
answered Jan 27 at 23:44
Calum GilhooleyCalum Gilhooley
5,119630
5,119630
1
$begingroup$
(I'm braced for a ticking-off from @fleablood, for indulging in unnecessary cheese-paring!) :)
$endgroup$
– Calum Gilhooley
Jan 27 at 23:47
add a comment |
1
$begingroup$
(I'm braced for a ticking-off from @fleablood, for indulging in unnecessary cheese-paring!) :)
$endgroup$
– Calum Gilhooley
Jan 27 at 23:47
1
1
$begingroup$
(I'm braced for a ticking-off from @fleablood, for indulging in unnecessary cheese-paring!) :)
$endgroup$
– Calum Gilhooley
Jan 27 at 23:47
$begingroup$
(I'm braced for a ticking-off from @fleablood, for indulging in unnecessary cheese-paring!) :)
$endgroup$
– Calum Gilhooley
Jan 27 at 23:47
add a comment |
$begingroup$
Note that$$frac14-frac{n-1}{4n+1}=frac5{4(4n+1)}$$and$$frac5{4(4n+1)}<varepsiloniff n>frac{5-4varepsilon}{16varepsilon}.$$
answered Jan 27 at 21:56
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Thanks. However, that doesn't show me a methodical way of reaching that conclusion.
$endgroup$
– jaja
Jan 28 at 0:00
add a comment |
$begingroup$
Note that$$frac14-frac{n-1}{4n+1}=frac5{4(4n+1)}$$and$$frac5{4(4n+1)}<varepsiloniff n>frac{5-4varepsilon}{16varepsilon}.$$
answered Jan 27 at 21:56
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
$begingroup$
Thanks. However, that doesn't show me a methodical way of reaching that conclusion.
$endgroup$
– jaja
Jan 28 at 0:00
add a comment |
$begingroup$
Note that$$frac14-frac{n-1}{4n+1}=frac5{4(4n+1)}$$and$$frac5{4(4n+1)}<varepsiloniff n>frac{5-4varepsilon}{16varepsilon}.$$
answered Jan 27 at 21:56
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
Note that$$frac14-frac{n-1}{4n+1}=frac5{4(4n+1)}$$and$$frac5{4(4n+1)}<varepsiloniff n>frac{5-4varepsilon}{16varepsilon}.$$
answered Jan 27 at 21:56
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jan 27 at 21:56
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jan 27 at 21:56
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jan 27 at 21:56
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
$begingroup$
Thanks. However, that doesn't show me a methodical way of reaching that conclusion.
$endgroup$
– jaja
Jan 28 at 0:00
add a comment |
$begingroup$
Thanks. However, that doesn't show me a methodical way of reaching that conclusion.
$endgroup$
– jaja
Jan 28 at 0:00
$begingroup$
Thanks. However, that doesn't show me a methodical way of reaching that conclusion.
$endgroup$
– jaja
Jan 28 at 0:00
$begingroup$
Thanks. However, that doesn't show me a methodical way of reaching that conclusion.
$endgroup$
– jaja
Jan 28 at 0:00
add a comment |
$begingroup$
Well, if $s = frac {n-1}{4n+1} in S; nin mathbb N$ then $frac {n-1}{4n+1} = frac {n+frac 14}{4n+1} - frac {frac 54}{4n+1} = frac 14 - frac {5}{16n + 4}< frac 14$.
So $frac 14$ is an upperbound of $S$.
If $k < frac 14$ and if $epsilon = frac 14 -k>0$.
$frac 5{16n + 4} < epsilon iff$
$frac 5{epsilon} < 16n + 4 iff$
$n > frac {frac 5epsilon -4}{16}$.
Such an $n$ can always be found and if we let $n > frac {frac 5epsilon -4}{16}$ we will have:
$k = frac 14 - epsilon < frac 14 - frac 5{16n+4} =frac {n-1}{4n+1}in S$.
So $k$ is not an upper bound of $S$.
And that is the definition of least upper bound. So $frac 14 =sup S$.
answered Jan 27 at 22:11
fleabloodfleablood
73.4k22791
$endgroup$
add a comment |
$begingroup$
Well, if $s = frac {n-1}{4n+1} in S; nin mathbb N$ then $frac {n-1}{4n+1} = frac {n+frac 14}{4n+1} - frac {frac 54}{4n+1} = frac 14 - frac {5}{16n + 4}< frac 14$.
So $frac 14$ is an upperbound of $S$.
If $k < frac 14$ and if $epsilon = frac 14 -k>0$.
$frac 5{16n + 4} < epsilon iff$
$frac 5{epsilon} < 16n + 4 iff$
$n > frac {frac 5epsilon -4}{16}$.
Such an $n$ can always be found and if we let $n > frac {frac 5epsilon -4}{16}$ we will have:
$k = frac 14 - epsilon < frac 14 - frac 5{16n+4} =frac {n-1}{4n+1}in S$.
So $k$ is not an upper bound of $S$.
And that is the definition of least upper bound. So $frac 14 =sup S$.
answered Jan 27 at 22:11
fleabloodfleablood
73.4k22791
$endgroup$
add a comment |
$begingroup$
Well, if $s = frac {n-1}{4n+1} in S; nin mathbb N$ then $frac {n-1}{4n+1} = frac {n+frac 14}{4n+1} - frac {frac 54}{4n+1} = frac 14 - frac {5}{16n + 4}< frac 14$.
So $frac 14$ is an upperbound of $S$.
If $k < frac 14$ and if $epsilon = frac 14 -k>0$.
$frac 5{16n + 4} < epsilon iff$
$frac 5{epsilon} < 16n + 4 iff$
$n > frac {frac 5epsilon -4}{16}$.
Such an $n$ can always be found and if we let $n > frac {frac 5epsilon -4}{16}$ we will have:
$k = frac 14 - epsilon < frac 14 - frac 5{16n+4} =frac {n-1}{4n+1}in S$.
So $k$ is not an upper bound of $S$.
And that is the definition of least upper bound. So $frac 14 =sup S$.
answered Jan 27 at 22:11
fleabloodfleablood
73.4k22791
$endgroup$
Well, if $s = frac {n-1}{4n+1} in S; nin mathbb N$ then $frac {n-1}{4n+1} = frac {n+frac 14}{4n+1} - frac {frac 54}{4n+1} = frac 14 - frac {5}{16n + 4}< frac 14$.
So $frac 14$ is an upperbound of $S$.
If $k < frac 14$ and if $epsilon = frac 14 -k>0$.
$frac 5{16n + 4} < epsilon iff$
$frac 5{epsilon} < 16n + 4 iff$
$n > frac {frac 5epsilon -4}{16}$.
Such an $n$ can always be found and if we let $n > frac {frac 5epsilon -4}{16}$ we will have:
$k = frac 14 - epsilon < frac 14 - frac 5{16n+4} =frac {n-1}{4n+1}in S$.
So $k$ is not an upper bound of $S$.
And that is the definition of least upper bound. So $frac 14 =sup S$.
answered Jan 27 at 22:11
fleabloodfleablood
73.4k22791
answered Jan 27 at 22:11
fleabloodfleablood
73.4k22791
answered Jan 27 at 22:11
fleabloodfleablood
73.4k22791
answered Jan 27 at 22:11
fleabloodfleablood
73.4k22791
73.4k22791
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090161%2fprove-that-the-supremum-of-this-set-is-frac14-s-fracn-14n1-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
