Mixing
Show that the series $sumlimits_{k=1}^{infty}frac{x^k}{k}$ does not converge uniformly on $(-1,1)$.
$begingroup$
Show that the series $sumlimits_{k=1}^{infty}dfrac{x^k}{k}$ does not converge uniformly on $(-1,1)$. (Hint: Show that the sequence of partial sums is not Cauchy in the sup-norm.)
Proof: Let $s_n = sumlimits_{k=1}^{n}dfrac{x^k}{k}$. We need to show that there exists $epsilon>0$ such that for all $Ninmathbb{N}$ and for $n geq N$ implies $|s_n-s_N|_{sup} geq epsilon$. As a matter of fact, I claim that this must be true for any $epsilon >0$ we choose. So let $epsilon >0$. But,
begin{align}
|s_n-s_N|_{sup} &= |sumlimits_{k=1}^{n}dfrac{x^k}{k} - sumlimits_{k=1}^{N}dfrac{x^k}{k}|_{sup} \
&= |sumlimits_{k=N+1}^{n}dfrac{x^k}{k}|_{sup} \
&= sumlimits_{k=N+1}^{n}dfrac{1}{k}
end{align}
which is a harmonic series and will diverge to infinity as $ntoinfty$. $blacksquare$
real-analysis sequences-and-series proof-verification uniform-convergence
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
asked Apr 21 '17 at 19:23
user3000482user3000482
761417
$endgroup$
add a comment |
$begingroup$
Show that the series $sumlimits_{k=1}^{infty}dfrac{x^k}{k}$ does not converge uniformly on $(-1,1)$. (Hint: Show that the sequence of partial sums is not Cauchy in the sup-norm.)
Proof: Let $s_n = sumlimits_{k=1}^{n}dfrac{x^k}{k}$. We need to show that there exists $epsilon>0$ such that for all $Ninmathbb{N}$ and for $n geq N$ implies $|s_n-s_N|_{sup} geq epsilon$. As a matter of fact, I claim that this must be true for any $epsilon >0$ we choose. So let $epsilon >0$. But,
begin{align}
|s_n-s_N|_{sup} &= |sumlimits_{k=1}^{n}dfrac{x^k}{k} - sumlimits_{k=1}^{N}dfrac{x^k}{k}|_{sup} \
&= |sumlimits_{k=N+1}^{n}dfrac{x^k}{k}|_{sup} \
&= sumlimits_{k=N+1}^{n}dfrac{1}{k}
end{align}
which is a harmonic series and will diverge to infinity as $ntoinfty$. $blacksquare$
real-analysis sequences-and-series proof-verification uniform-convergence
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
asked Apr 21 '17 at 19:23
user3000482user3000482
761417
$endgroup$
$begingroup$
Your proof is fine.
$endgroup$
– Starfall
Apr 21 '17 at 19:25
$begingroup$
Possible duplicate of A dubious proof using Weierstrass-M test for $sum^n_{k=1}frac{x^k}{k}$
$endgroup$
– Nosrati
Apr 21 '17 at 19:28
$begingroup$
you could calculate the value of series explicitly by integrating the simple geometric series ... then you will see that it is unbounded on $(-1,1)$.
$endgroup$
– Red shoes
Apr 21 '17 at 19:40
add a comment |
$begingroup$
Show that the series $sumlimits_{k=1}^{infty}dfrac{x^k}{k}$ does not converge uniformly on $(-1,1)$. (Hint: Show that the sequence of partial sums is not Cauchy in the sup-norm.)
Proof: Let $s_n = sumlimits_{k=1}^{n}dfrac{x^k}{k}$. We need to show that there exists $epsilon>0$ such that for all $Ninmathbb{N}$ and for $n geq N$ implies $|s_n-s_N|_{sup} geq epsilon$. As a matter of fact, I claim that this must be true for any $epsilon >0$ we choose. So let $epsilon >0$. But,
begin{align}
|s_n-s_N|_{sup} &= |sumlimits_{k=1}^{n}dfrac{x^k}{k} - sumlimits_{k=1}^{N}dfrac{x^k}{k}|_{sup} \
&= |sumlimits_{k=N+1}^{n}dfrac{x^k}{k}|_{sup} \
&= sumlimits_{k=N+1}^{n}dfrac{1}{k}
end{align}
which is a harmonic series and will diverge to infinity as $ntoinfty$. $blacksquare$
real-analysis sequences-and-series proof-verification uniform-convergence
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
asked Apr 21 '17 at 19:23
user3000482user3000482
761417
$endgroup$
Show that the series $sumlimits_{k=1}^{infty}dfrac{x^k}{k}$ does not converge uniformly on $(-1,1)$. (Hint: Show that the sequence of partial sums is not Cauchy in the sup-norm.)
Proof: Let $s_n = sumlimits_{k=1}^{n}dfrac{x^k}{k}$. We need to show that there exists $epsilon>0$ such that for all $Ninmathbb{N}$ and for $n geq N$ implies $|s_n-s_N|_{sup} geq epsilon$. As a matter of fact, I claim that this must be true for any $epsilon >0$ we choose. So let $epsilon >0$. But,
begin{align}
|s_n-s_N|_{sup} &= |sumlimits_{k=1}^{n}dfrac{x^k}{k} - sumlimits_{k=1}^{N}dfrac{x^k}{k}|_{sup} \
&= |sumlimits_{k=N+1}^{n}dfrac{x^k}{k}|_{sup} \
&= sumlimits_{k=N+1}^{n}dfrac{1}{k}
end{align}
which is a harmonic series and will diverge to infinity as $ntoinfty$. $blacksquare$
real-analysis sequences-and-series proof-verification uniform-convergence
real-analysis sequences-and-series proof-verification uniform-convergence
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
asked Apr 21 '17 at 19:23
user3000482user3000482
761417
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
asked Apr 21 '17 at 19:23
user3000482user3000482
761417
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
1,8402520
asked Apr 21 '17 at 19:23
user3000482user3000482
761417
asked Apr 21 '17 at 19:23
user3000482user3000482
761417
asked Apr 21 '17 at 19:23
user3000482user3000482
761417
761417
$begingroup$
Your proof is fine.
$endgroup$
– Starfall
Apr 21 '17 at 19:25
$begingroup$
Possible duplicate of A dubious proof using Weierstrass-M test for $sum^n_{k=1}frac{x^k}{k}$
$endgroup$
– Nosrati
Apr 21 '17 at 19:28
$begingroup$
you could calculate the value of series explicitly by integrating the simple geometric series ... then you will see that it is unbounded on $(-1,1)$.
$endgroup$
– Red shoes
Apr 21 '17 at 19:40
add a comment |
$begingroup$
Your proof is fine.
$endgroup$
– Starfall
Apr 21 '17 at 19:25
$begingroup$
Possible duplicate of A dubious proof using Weierstrass-M test for $sum^n_{k=1}frac{x^k}{k}$
$endgroup$
– Nosrati
Apr 21 '17 at 19:28
$begingroup$
you could calculate the value of series explicitly by integrating the simple geometric series ... then you will see that it is unbounded on $(-1,1)$.
$endgroup$
– Red shoes
Apr 21 '17 at 19:40
$begingroup$
Your proof is fine.
$endgroup$
– Starfall
Apr 21 '17 at 19:25
$begingroup$
Your proof is fine.
$endgroup$
– Starfall
Apr 21 '17 at 19:25
$begingroup$
Possible duplicate of A dubious proof using Weierstrass-M test for $sum^n_{k=1}frac{x^k}{k}$
$endgroup$
– Nosrati
Apr 21 '17 at 19:28
$begingroup$
Possible duplicate of A dubious proof using Weierstrass-M test for $sum^n_{k=1}frac{x^k}{k}$
$endgroup$
– Nosrati
Apr 21 '17 at 19:28
$begingroup$
you could calculate the value of series explicitly by integrating the simple geometric series ... then you will see that it is unbounded on $(-1,1)$.
$endgroup$
– Red shoes
Apr 21 '17 at 19:40
$begingroup$
you could calculate the value of series explicitly by integrating the simple geometric series ... then you will see that it is unbounded on $(-1,1)$.
$endgroup$
– Red shoes
Apr 21 '17 at 19:40
add a comment |
0
active
oldest
votes
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%2f2245493%2fshow-that-the-series-sum-limits-k-1-infty-fracxkk-does-not-converge%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f2245493%2fshow-that-the-series-sum-limits-k-1-infty-fracxkk-does-not-converge%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
$begingroup$
Your proof is fine.
$endgroup$
– Starfall
Apr 21 '17 at 19:25
$begingroup$
Possible duplicate of A dubious proof using Weierstrass-M test for $sum^n_{k=1}frac{x^k}{k}$
$endgroup$
– Nosrati
Apr 21 '17 at 19:28
$begingroup$
you could calculate the value of series explicitly by integrating the simple geometric series ... then you will see that it is unbounded on $(-1,1)$.
$endgroup$
– Red shoes
Apr 21 '17 at 19:40