proofverification of $lim_nrightarrowinftysqrt[n]n+s=1$ where $sge 0$ and request for an easier proof The 2019 Stack Overflow Developer Survey Results Are InProof that $lim_nrightarrow infty sqrt[n]n=1$Find $lim_n rightarrow inftysqrtn(A_n+1 − A_n)$ where $A_n = frac1n(a_1 + a_2 + cdots + a_n)$troubles proving $lim_nrightarrowinftysqrt[n]n=1$How to prove $lim_n to inftya_n=1 rightarrow lim_n to inftysqrt[n] a_n=1$find $lim_n rightarrow inftysqrtn(A_n+1 − A_n)$Prove $lim_nrightarrowinfty(a_n + b_n) = a + b$ if $lim_nrightarrowinftya_n=a$ and $lim_nrightarrowinftyb_n=b$When does equality for $x_n < y_n Rightarrow lim_n rightarrow infty x_n leq lim_n rightarrow infty y_n$ hold?If $lim_n rightarrow infty a_n = L > 0$. Prove. $lim_n rightarrow infty sqrta_n = sqrtL$ convergesShow that $limlimits_nrightarrowinfty(nb^n)=0$ for $0<b<1$Find the limit $lim_ntoinftyfracx_n + x_n^2 + cdots + x_n^k - kx_n - 1$ given $x_n ne 1$ and $lim x_n = 1$
Who coined the term "madman theory"?
Statement true because not provable
Write faster on AT24C32
FPGA - DIY Programming
Is this app Icon Browser Safe/Legit?
What does ひと匙 mean in this manga and has it been used colloquially?
Building a conditional check constraint
Why is the maximum length of OpenWrt’s root password 8 characters?
Is flight data recorder erased after every flight?
Protecting Dualbooting Windows from dangerous code (like rm -rf)
What is the meaning of the verb "bear" in this context?
Can someone be penalized for an "unlawful" act if no penalty is specified?
How to deal with fear of taking dependencies
What to do when moving next to a bird sanctuary with a loosely-domesticated cat?
Why do UK politicians seemingly ignore opinion polls on Brexit?
Does the shape of a die affect the probability of a number being rolled?
Are there incongruent pythagorean triangles with the same perimeter and same area?
Falsification in Math vs Science
How to notate time signature switching consistently every measure
Sci-fi book where a human is taken from Earth to help man an alien ship in a fight against other aliens and rises through the ranks to command
One word riddle: Vowel in the middle
How are Package `Private` variables accessed?
Is "plugging out" electronic devices an American expression?
Why was M87 targetted for the Event Horizon Telescope instead of Sagittarius A*?
proofverification of $lim_nrightarrowinftysqrt[n]n+s=1$ where $sge 0$ and request for an easier proof
The 2019 Stack Overflow Developer Survey Results Are InProof that $lim_nrightarrow infty sqrt[n]n=1$Find $lim_n rightarrow inftysqrtn(A_n+1 − A_n)$ where $A_n = frac1n(a_1 + a_2 + cdots + a_n)$troubles proving $lim_nrightarrowinftysqrt[n]n=1$How to prove $lim_n to inftya_n=1 rightarrow lim_n to inftysqrt[n] a_n=1$find $lim_n rightarrow inftysqrtn(A_n+1 − A_n)$Prove $lim_nrightarrowinfty(a_n + b_n) = a + b$ if $lim_nrightarrowinftya_n=a$ and $lim_nrightarrowinftyb_n=b$When does equality for $x_n < y_n Rightarrow lim_n rightarrow infty x_n leq lim_n rightarrow infty y_n$ hold?If $lim_n rightarrow infty a_n = L > 0$. Prove. $lim_n rightarrow infty sqrta_n = sqrtL$ convergesShow that $limlimits_nrightarrowinfty(nb^n)=0$ for $0<b<1$Find the limit $lim_ntoinftyfracx_n + x_n^2 + cdots + x_n^k - kx_n - 1$ given $x_n ne 1$ and $lim x_n = 1$
$begingroup$
I have basically copied the original proof that $lim_nrightarrowinftysqrt[n]n=1$ and I hope that there are no mistakes:
$x_n=sqrt[n]n+s-1geq0$
$n+s=(x_n+1)^ngeq1+binomn2x_n^2=1+fracn(n-1)2x_n^2$
Therefore $x_nleqsqrtfrac2(n+s-1)n(n-1)$
And therefore $0leqsqrt[n]n+s-1=x_nleq sqrtfrac2(n+s-1)n(n-1)$
From now on I have stoppped to copy the original proof
$lim$ stands for $lim_nrightarrow infty$
And because $limfrac2(n-1)2(n-1)+2s=lim 1-frac2s2(n-1)+2s=1$ And the rule if $lim a_n=a>0$ then $lim sqrt[k]a_n=sqrt[k]a$
I can say (I am not sure if I am really allowed to do that, some verification would be appreciated)
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2(n+s-1)n(n-1)$$
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2(n+s-1)n(n-1)cdot 1$$
This is the step where I am asking for verification
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2(n-1)+2sn(n-1)cdot sqrtfrac2(n-1)2(n-1)+2s$$
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2n=0$$
Is the proof alright? I hope there is an easier one
real-analysis sequences-and-series proof-verification
asked Mar 30 at 16:15
New2MathNew2Math
18315
$endgroup$
add a comment |
$begingroup$
I have basically copied the original proof that $lim_nrightarrowinftysqrt[n]n=1$ and I hope that there are no mistakes:
$x_n=sqrt[n]n+s-1geq0$
$n+s=(x_n+1)^ngeq1+binomn2x_n^2=1+fracn(n-1)2x_n^2$
Therefore $x_nleqsqrtfrac2(n+s-1)n(n-1)$
And therefore $0leqsqrt[n]n+s-1=x_nleq sqrtfrac2(n+s-1)n(n-1)$
From now on I have stoppped to copy the original proof
$lim$ stands for $lim_nrightarrow infty$
And because $limfrac2(n-1)2(n-1)+2s=lim 1-frac2s2(n-1)+2s=1$ And the rule if $lim a_n=a>0$ then $lim sqrt[k]a_n=sqrt[k]a$
I can say (I am not sure if I am really allowed to do that, some verification would be appreciated)
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2(n+s-1)n(n-1)$$
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2(n+s-1)n(n-1)cdot 1$$
This is the step where I am asking for verification
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2(n-1)+2sn(n-1)cdot sqrtfrac2(n-1)2(n-1)+2s$$
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2n=0$$
Is the proof alright? I hope there is an easier one
real-analysis sequences-and-series proof-verification
asked Mar 30 at 16:15
New2MathNew2Math
18315
$endgroup$
1
$begingroup$
For an easier proof, notice $sqrt[n]nlesqrt[n]n+slesqrt[n]2n$ for sufficiently large $n$ and then squeeze.
$endgroup$
– Thorgott
Mar 30 at 16:21
$begingroup$
@Thorgott Why not an official answer?
$endgroup$
– Paul Frost
Mar 30 at 16:23
$begingroup$
Why the restriction on $s$?
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 16:24
$begingroup$
If s is negative the root does not exist (yet) for $sqrt-1$
$endgroup$
– New2Math
Mar 30 at 16:25
$begingroup$
@New2Math, then $sge 0$ is enough. $sinBbb N$ isn't required.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 16:27
add a comment |
$begingroup$
I have basically copied the original proof that $lim_nrightarrowinftysqrt[n]n=1$ and I hope that there are no mistakes:
$x_n=sqrt[n]n+s-1geq0$
$n+s=(x_n+1)^ngeq1+binomn2x_n^2=1+fracn(n-1)2x_n^2$
Therefore $x_nleqsqrtfrac2(n+s-1)n(n-1)$
And therefore $0leqsqrt[n]n+s-1=x_nleq sqrtfrac2(n+s-1)n(n-1)$
From now on I have stoppped to copy the original proof
$lim$ stands for $lim_nrightarrow infty$
And because $limfrac2(n-1)2(n-1)+2s=lim 1-frac2s2(n-1)+2s=1$ And the rule if $lim a_n=a>0$ then $lim sqrt[k]a_n=sqrt[k]a$
I can say (I am not sure if I am really allowed to do that, some verification would be appreciated)
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2(n+s-1)n(n-1)$$
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2(n+s-1)n(n-1)cdot 1$$
This is the step where I am asking for verification
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2(n-1)+2sn(n-1)cdot sqrtfrac2(n-1)2(n-1)+2s$$
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2n=0$$
Is the proof alright? I hope there is an easier one
real-analysis sequences-and-series proof-verification
asked Mar 30 at 16:15
New2MathNew2Math
18315
$endgroup$
I have basically copied the original proof that $lim_nrightarrowinftysqrt[n]n=1$ and I hope that there are no mistakes:
$x_n=sqrt[n]n+s-1geq0$
$n+s=(x_n+1)^ngeq1+binomn2x_n^2=1+fracn(n-1)2x_n^2$
Therefore $x_nleqsqrtfrac2(n+s-1)n(n-1)$
And therefore $0leqsqrt[n]n+s-1=x_nleq sqrtfrac2(n+s-1)n(n-1)$
From now on I have stoppped to copy the original proof
$lim$ stands for $lim_nrightarrow infty$
And because $limfrac2(n-1)2(n-1)+2s=lim 1-frac2s2(n-1)+2s=1$ And the rule if $lim a_n=a>0$ then $lim sqrt[k]a_n=sqrt[k]a$
I can say (I am not sure if I am really allowed to do that, some verification would be appreciated)
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2(n+s-1)n(n-1)$$
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2(n+s-1)n(n-1)cdot 1$$
This is the step where I am asking for verification
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2(n-1)+2sn(n-1)cdot sqrtfrac2(n-1)2(n-1)+2s$$
$$lim 0leq limsqrt[n]n+s-1=lim x_nleq lim sqrtfrac2n=0$$
Is the proof alright? I hope there is an easier one
real-analysis sequences-and-series proof-verification
real-analysis sequences-and-series proof-verification
asked Mar 30 at 16:15
New2MathNew2Math
18315
asked Mar 30 at 16:15
New2MathNew2Math
18315
edited Mar 30 at 16:30
New2Math
asked Mar 30 at 16:15
New2MathNew2Math
18315
asked Mar 30 at 16:15
New2MathNew2Math
18315
asked Mar 30 at 16:15
New2MathNew2Math
18315
18315
1
$begingroup$
For an easier proof, notice $sqrt[n]nlesqrt[n]n+slesqrt[n]2n$ for sufficiently large $n$ and then squeeze.
$endgroup$
– Thorgott
Mar 30 at 16:21
$begingroup$
@Thorgott Why not an official answer?
$endgroup$
– Paul Frost
Mar 30 at 16:23
$begingroup$
Why the restriction on $s$?
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 16:24
$begingroup$
If s is negative the root does not exist (yet) for $sqrt-1$
$endgroup$
– New2Math
Mar 30 at 16:25
$begingroup$
@New2Math, then $sge 0$ is enough. $sinBbb N$ isn't required.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 16:27
add a comment |
1
$begingroup$
For an easier proof, notice $sqrt[n]nlesqrt[n]n+slesqrt[n]2n$ for sufficiently large $n$ and then squeeze.
$endgroup$
– Thorgott
Mar 30 at 16:21
$begingroup$
@Thorgott Why not an official answer?
$endgroup$
– Paul Frost
Mar 30 at 16:23
$begingroup$
Why the restriction on $s$?
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 16:24
$begingroup$
If s is negative the root does not exist (yet) for $sqrt-1$
$endgroup$
– New2Math
Mar 30 at 16:25
$begingroup$
@New2Math, then $sge 0$ is enough. $sinBbb N$ isn't required.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 16:27
1
1
$begingroup$
For an easier proof, notice $sqrt[n]nlesqrt[n]n+slesqrt[n]2n$ for sufficiently large $n$ and then squeeze.
$endgroup$
– Thorgott
Mar 30 at 16:21
$begingroup$
For an easier proof, notice $sqrt[n]nlesqrt[n]n+slesqrt[n]2n$ for sufficiently large $n$ and then squeeze.
$endgroup$
– Thorgott
Mar 30 at 16:21
$begingroup$
@Thorgott Why not an official answer?
$endgroup$
– Paul Frost
Mar 30 at 16:23
$begingroup$
@Thorgott Why not an official answer?
$endgroup$
– Paul Frost
Mar 30 at 16:23
$begingroup$
Why the restriction on $s$?
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 16:24
$begingroup$
Why the restriction on $s$?
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 16:24
$begingroup$
If s is negative the root does not exist (yet) for $sqrt-1$
$endgroup$
– New2Math
Mar 30 at 16:25
$begingroup$
If s is negative the root does not exist (yet) for $sqrt-1$
$endgroup$
– New2Math
Mar 30 at 16:25
$begingroup$
@New2Math, then $sge 0$ is enough. $sinBbb N$ isn't required.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 16:27
$begingroup$
@New2Math, then $sge 0$ is enough. $sinBbb N$ isn't required.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 16:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Easier proof:
$$lim_ntoinftylog(sqrt[n]n+s) = lim_ntoinftyfraclog(n+s)n = 0.$$
answered Mar 30 at 16:27
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
$endgroup$
add a comment |
$begingroup$
I think an easier one, goes like this:$$lim_nto inftysqrt[n]an=lim_nto inftysqrt[n]asqrt[n]n\=lim_nto infty1cdotsqrt[n]n\=lim_nto inftysqrt[n]n$$for $a>0$. Also$$lim_nto inftysqrt[n]n=lim_uto infty e^ue^-u=e^0=1$$therefore$$lim_nto inftysqrt[n]an=1$$ and hence$$1=lim_nto inftysqrt[n]nlelim_nto inftysqrt[n]n+sle lim_nto inftysqrt[n]2n=1$$which implies that$$lim_nto inftysqrt[n]n+s=1$$
answered Apr 3 at 17:22
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
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%2f3168462%2fproofverification-of-lim-n-rightarrow-infty-sqrtnns-1-where-s-ge-0-a%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
$begingroup$
Easier proof:
$$lim_ntoinftylog(sqrt[n]n+s) = lim_ntoinftyfraclog(n+s)n = 0.$$
answered Mar 30 at 16:27
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
$endgroup$
add a comment |
$begingroup$
Easier proof:
$$lim_ntoinftylog(sqrt[n]n+s) = lim_ntoinftyfraclog(n+s)n = 0.$$
answered Mar 30 at 16:27
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
$endgroup$
add a comment |
$begingroup$
Easier proof:
$$lim_ntoinftylog(sqrt[n]n+s) = lim_ntoinftyfraclog(n+s)n = 0.$$
answered Mar 30 at 16:27
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
$endgroup$
Easier proof:
$$lim_ntoinftylog(sqrt[n]n+s) = lim_ntoinftyfraclog(n+s)n = 0.$$
answered Mar 30 at 16:27
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
answered Mar 30 at 16:27
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
answered Mar 30 at 16:27
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
answered Mar 30 at 16:27
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
35.4k42972
add a comment |
add a comment |
$begingroup$
I think an easier one, goes like this:$$lim_nto inftysqrt[n]an=lim_nto inftysqrt[n]asqrt[n]n\=lim_nto infty1cdotsqrt[n]n\=lim_nto inftysqrt[n]n$$for $a>0$. Also$$lim_nto inftysqrt[n]n=lim_uto infty e^ue^-u=e^0=1$$therefore$$lim_nto inftysqrt[n]an=1$$ and hence$$1=lim_nto inftysqrt[n]nlelim_nto inftysqrt[n]n+sle lim_nto inftysqrt[n]2n=1$$which implies that$$lim_nto inftysqrt[n]n+s=1$$
answered Apr 3 at 17:22
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
I think an easier one, goes like this:$$lim_nto inftysqrt[n]an=lim_nto inftysqrt[n]asqrt[n]n\=lim_nto infty1cdotsqrt[n]n\=lim_nto inftysqrt[n]n$$for $a>0$. Also$$lim_nto inftysqrt[n]n=lim_uto infty e^ue^-u=e^0=1$$therefore$$lim_nto inftysqrt[n]an=1$$ and hence$$1=lim_nto inftysqrt[n]nlelim_nto inftysqrt[n]n+sle lim_nto inftysqrt[n]2n=1$$which implies that$$lim_nto inftysqrt[n]n+s=1$$
answered Apr 3 at 17:22
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
I think an easier one, goes like this:$$lim_nto inftysqrt[n]an=lim_nto inftysqrt[n]asqrt[n]n\=lim_nto infty1cdotsqrt[n]n\=lim_nto inftysqrt[n]n$$for $a>0$. Also$$lim_nto inftysqrt[n]n=lim_uto infty e^ue^-u=e^0=1$$therefore$$lim_nto inftysqrt[n]an=1$$ and hence$$1=lim_nto inftysqrt[n]nlelim_nto inftysqrt[n]n+sle lim_nto inftysqrt[n]2n=1$$which implies that$$lim_nto inftysqrt[n]n+s=1$$
answered Apr 3 at 17:22
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
I think an easier one, goes like this:$$lim_nto inftysqrt[n]an=lim_nto inftysqrt[n]asqrt[n]n\=lim_nto infty1cdotsqrt[n]n\=lim_nto inftysqrt[n]n$$for $a>0$. Also$$lim_nto inftysqrt[n]n=lim_uto infty e^ue^-u=e^0=1$$therefore$$lim_nto inftysqrt[n]an=1$$ and hence$$1=lim_nto inftysqrt[n]nlelim_nto inftysqrt[n]n+sle lim_nto inftysqrt[n]2n=1$$which implies that$$lim_nto inftysqrt[n]n+s=1$$
answered Apr 3 at 17:22
Mostafa AyazMostafa Ayaz
18.1k31040
answered Apr 3 at 17:22
Mostafa AyazMostafa Ayaz
18.1k31040
answered Apr 3 at 17:22
Mostafa AyazMostafa Ayaz
18.1k31040
answered Apr 3 at 17:22
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
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%2f3168462%2fproofverification-of-lim-n-rightarrow-infty-sqrtnns-1-where-s-ge-0-a%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
1
$begingroup$
For an easier proof, notice $sqrt[n]nlesqrt[n]n+slesqrt[n]2n$ for sufficiently large $n$ and then squeeze.
$endgroup$
– Thorgott
Mar 30 at 16:21
$begingroup$
@Thorgott Why not an official answer?
$endgroup$
– Paul Frost
Mar 30 at 16:23
$begingroup$
Why the restriction on $s$?
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 16:24
$begingroup$
If s is negative the root does not exist (yet) for $sqrt-1$
$endgroup$
– New2Math
Mar 30 at 16:25
$begingroup$
@New2Math, then $sge 0$ is enough. $sinBbb N$ isn't required.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 16:27