prove that $f_n(x) = frac{n^2x}{1 + n^2x^2log n}$ converge but not uniform












2












$begingroup$



I have to prove that the sequence of functions
$$
f_n(x) = frac{n^2x}{1 + n^2x^2log n},
$$ converges.




So I think that I can just show that $lim_{n to infty} (f_n) = 0$ pointwise. But to show that the sequence does not converge uniformly, I need to show that $f_n$ has a point of maximum $x = frac{1}{n sqrt{log n}}$ and at this point, the limit is equal to infinity. Is that correct?



And finally I need to show that $forall alpha > 0$ in the interval $mathbb{R} setminus (-alpha,+alpha)$, the sequence $f_n$ converges uniformly, but I don't have ideas how to do it.






real-analysis sequences-and-series analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    You do not need to prove that the limit is infinity, you just need to prove that the sequence does not go to zero.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 21:39










  • $begingroup$
    Okay, I got it, for example $f_n(x) = x^n(1-x^n)$. When $x = sqrt[n]{1/2}$ we have $f_n = 1/4$, thus is not uniformly convergent, correct?
    $endgroup$
    – Matheus Fachini
    Dec 15 '18 at 21:45






  • 1




    $begingroup$
    Yes, because the max-norm is always at least $1/4$ so cannot go to zero.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:23
















2












$begingroup$



I have to prove that the sequence of functions
$$
f_n(x) = frac{n^2x}{1 + n^2x^2log n},
$$ converges.




So I think that I can just show that $lim_{n to infty} (f_n) = 0$ pointwise. But to show that the sequence does not converge uniformly, I need to show that $f_n$ has a point of maximum $x = frac{1}{n sqrt{log n}}$ and at this point, the limit is equal to infinity. Is that correct?



And finally I need to show that $forall alpha > 0$ in the interval $mathbb{R} setminus (-alpha,+alpha)$, the sequence $f_n$ converges uniformly, but I don't have ideas how to do it.






real-analysis sequences-and-series analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    You do not need to prove that the limit is infinity, you just need to prove that the sequence does not go to zero.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 21:39










  • $begingroup$
    Okay, I got it, for example $f_n(x) = x^n(1-x^n)$. When $x = sqrt[n]{1/2}$ we have $f_n = 1/4$, thus is not uniformly convergent, correct?
    $endgroup$
    – Matheus Fachini
    Dec 15 '18 at 21:45






  • 1




    $begingroup$
    Yes, because the max-norm is always at least $1/4$ so cannot go to zero.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:23














2












2








2


1



$begingroup$



I have to prove that the sequence of functions
$$
f_n(x) = frac{n^2x}{1 + n^2x^2log n},
$$ converges.




So I think that I can just show that $lim_{n to infty} (f_n) = 0$ pointwise. But to show that the sequence does not converge uniformly, I need to show that $f_n$ has a point of maximum $x = frac{1}{n sqrt{log n}}$ and at this point, the limit is equal to infinity. Is that correct?



And finally I need to show that $forall alpha > 0$ in the interval $mathbb{R} setminus (-alpha,+alpha)$, the sequence $f_n$ converges uniformly, but I don't have ideas how to do it.






real-analysis sequences-and-series analysis







share|cite|improve this question











$endgroup$





I have to prove that the sequence of functions
$$
f_n(x) = frac{n^2x}{1 + n^2x^2log n},
$$ converges.




So I think that I can just show that $lim_{n to infty} (f_n) = 0$ pointwise. But to show that the sequence does not converge uniformly, I need to show that $f_n$ has a point of maximum $x = frac{1}{n sqrt{log n}}$ and at this point, the limit is equal to infinity. Is that correct?



And finally I need to show that $forall alpha > 0$ in the interval $mathbb{R} setminus (-alpha,+alpha)$, the sequence $f_n$ converges uniformly, but I don't have ideas how to do it.






real-analysis sequences-and-series analysis




real-analysis sequences-and-series analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 21:33







user587192

















asked Dec 15 '18 at 21:24









Matheus FachiniMatheus Fachini

859




859












  • $begingroup$
    You do not need to prove that the limit is infinity, you just need to prove that the sequence does not go to zero.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 21:39










  • $begingroup$
    Okay, I got it, for example $f_n(x) = x^n(1-x^n)$. When $x = sqrt[n]{1/2}$ we have $f_n = 1/4$, thus is not uniformly convergent, correct?
    $endgroup$
    – Matheus Fachini
    Dec 15 '18 at 21:45






  • 1




    $begingroup$
    Yes, because the max-norm is always at least $1/4$ so cannot go to zero.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:23


















  • $begingroup$
    You do not need to prove that the limit is infinity, you just need to prove that the sequence does not go to zero.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 21:39










  • $begingroup$
    Okay, I got it, for example $f_n(x) = x^n(1-x^n)$. When $x = sqrt[n]{1/2}$ we have $f_n = 1/4$, thus is not uniformly convergent, correct?
    $endgroup$
    – Matheus Fachini
    Dec 15 '18 at 21:45






  • 1




    $begingroup$
    Yes, because the max-norm is always at least $1/4$ so cannot go to zero.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:23
















$begingroup$
You do not need to prove that the limit is infinity, you just need to prove that the sequence does not go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 21:39




$begingroup$
You do not need to prove that the limit is infinity, you just need to prove that the sequence does not go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 21:39












$begingroup$
Okay, I got it, for example $f_n(x) = x^n(1-x^n)$. When $x = sqrt[n]{1/2}$ we have $f_n = 1/4$, thus is not uniformly convergent, correct?
$endgroup$
– Matheus Fachini
Dec 15 '18 at 21:45




$begingroup$
Okay, I got it, for example $f_n(x) = x^n(1-x^n)$. When $x = sqrt[n]{1/2}$ we have $f_n = 1/4$, thus is not uniformly convergent, correct?
$endgroup$
– Matheus Fachini
Dec 15 '18 at 21:45




1




1




$begingroup$
Yes, because the max-norm is always at least $1/4$ so cannot go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 22:23




$begingroup$
Yes, because the max-norm is always at least $1/4$ so cannot go to zero.
$endgroup$
– Mindlack
Dec 15 '18 at 22:23










1 Answer
1






active

oldest

votes


















2












$begingroup$

$1). $ If $x=0$ the sequence clearly converges to $0$. If not, then $|f_n(x)|le frac{1}{xlog n}to 0$ as $nto infty$ so $(f_n)$ is pointwise convergent.



$2). $ The trouble will be near zero, so take $x_n=1/n$. Then $x_nto 0$ whereas $f_n(x_n)to infty$, and so $(f_n)$ is not uniformly convergent.



$3). $ $f_n$ is odd for each $nge 1$ so it suffices to check $[alpha,infty).$ In this case,$ |f_n(x)|le frac{1}{alphalog n}$ so $(f_n)$ is uniformly bounded, and so converges uniformly.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Because $f_n(1/n)=n/(1+ln n)$
    $endgroup$
    – Matematleta
    Dec 16 '18 at 2:39










  • $begingroup$
    I had misread the numerator as $n^2x^2$.
    $endgroup$
    – Mark Viola
    Dec 16 '18 at 14:39











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%2f3041979%2fprove-that-f-nx-fracn2x1-n2x2-log-n-converge-but-not-uniform%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

$1). $ If $x=0$ the sequence clearly converges to $0$. If not, then $|f_n(x)|le frac{1}{xlog n}to 0$ as $nto infty$ so $(f_n)$ is pointwise convergent.



$2). $ The trouble will be near zero, so take $x_n=1/n$. Then $x_nto 0$ whereas $f_n(x_n)to infty$, and so $(f_n)$ is not uniformly convergent.



$3). $ $f_n$ is odd for each $nge 1$ so it suffices to check $[alpha,infty).$ In this case,$ |f_n(x)|le frac{1}{alphalog n}$ so $(f_n)$ is uniformly bounded, and so converges uniformly.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Because $f_n(1/n)=n/(1+ln n)$
    $endgroup$
    – Matematleta
    Dec 16 '18 at 2:39










  • $begingroup$
    I had misread the numerator as $n^2x^2$.
    $endgroup$
    – Mark Viola
    Dec 16 '18 at 14:39
















2












$begingroup$

$1). $ If $x=0$ the sequence clearly converges to $0$. If not, then $|f_n(x)|le frac{1}{xlog n}to 0$ as $nto infty$ so $(f_n)$ is pointwise convergent.



$2). $ The trouble will be near zero, so take $x_n=1/n$. Then $x_nto 0$ whereas $f_n(x_n)to infty$, and so $(f_n)$ is not uniformly convergent.



$3). $ $f_n$ is odd for each $nge 1$ so it suffices to check $[alpha,infty).$ In this case,$ |f_n(x)|le frac{1}{alphalog n}$ so $(f_n)$ is uniformly bounded, and so converges uniformly.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Because $f_n(1/n)=n/(1+ln n)$
    $endgroup$
    – Matematleta
    Dec 16 '18 at 2:39










  • $begingroup$
    I had misread the numerator as $n^2x^2$.
    $endgroup$
    – Mark Viola
    Dec 16 '18 at 14:39














2












2








2





$begingroup$

$1). $ If $x=0$ the sequence clearly converges to $0$. If not, then $|f_n(x)|le frac{1}{xlog n}to 0$ as $nto infty$ so $(f_n)$ is pointwise convergent.



$2). $ The trouble will be near zero, so take $x_n=1/n$. Then $x_nto 0$ whereas $f_n(x_n)to infty$, and so $(f_n)$ is not uniformly convergent.



$3). $ $f_n$ is odd for each $nge 1$ so it suffices to check $[alpha,infty).$ In this case,$ |f_n(x)|le frac{1}{alphalog n}$ so $(f_n)$ is uniformly bounded, and so converges uniformly.






share|cite|improve this answer











$endgroup$



$1). $ If $x=0$ the sequence clearly converges to $0$. If not, then $|f_n(x)|le frac{1}{xlog n}to 0$ as $nto infty$ so $(f_n)$ is pointwise convergent.



$2). $ The trouble will be near zero, so take $x_n=1/n$. Then $x_nto 0$ whereas $f_n(x_n)to infty$, and so $(f_n)$ is not uniformly convergent.



$3). $ $f_n$ is odd for each $nge 1$ so it suffices to check $[alpha,infty).$ In this case,$ |f_n(x)|le frac{1}{alphalog n}$ so $(f_n)$ is uniformly bounded, and so converges uniformly.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 4:51

























answered Dec 15 '18 at 22:24









MatematletaMatematleta

11.5k2920




11.5k2920












  • $begingroup$
    Because $f_n(1/n)=n/(1+ln n)$
    $endgroup$
    – Matematleta
    Dec 16 '18 at 2:39










  • $begingroup$
    I had misread the numerator as $n^2x^2$.
    $endgroup$
    – Mark Viola
    Dec 16 '18 at 14:39


















  • $begingroup$
    Because $f_n(1/n)=n/(1+ln n)$
    $endgroup$
    – Matematleta
    Dec 16 '18 at 2:39










  • $begingroup$
    I had misread the numerator as $n^2x^2$.
    $endgroup$
    – Mark Viola
    Dec 16 '18 at 14:39
















$begingroup$
Because $f_n(1/n)=n/(1+ln n)$
$endgroup$
– Matematleta
Dec 16 '18 at 2:39




$begingroup$
Because $f_n(1/n)=n/(1+ln n)$
$endgroup$
– Matematleta
Dec 16 '18 at 2:39












$begingroup$
I had misread the numerator as $n^2x^2$.
$endgroup$
– Mark Viola
Dec 16 '18 at 14:39




$begingroup$
I had misread the numerator as $n^2x^2$.
$endgroup$
– Mark Viola
Dec 16 '18 at 14:39


















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%2f3041979%2fprove-that-f-nx-fracn2x1-n2x2-log-n-converge-but-not-uniform%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

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more