Questions about the powerseries $sum_{n=0}^{infty}frac{1}{n^d}z^n,dinmathbb{R},zinmathbb{C}$
$begingroup$
The convergenceradius is $1$ that means if $|z|= 1$ then the convergencebehaviour depends on $d$.
I have several questions about this issue:
1.Why is $sum_{n=0}^{infty}frac{1}{n^d}$ absolutely convergent for $d>1$ and $|z|=1$
2.Why is the implication true:
series absolutely divergent $Rightarrow$ series is divergent
3.Why is $sum_{n=0}^{infty}frac{1}{n^d}z^n$ divergent for $dleq 0$ and $|z|=1$
4.Why does the convergencebehaviour depends on $zin{ainmathbb{C}||a|=1}$ for $0<dleq 1$?
I know that if $d$ is for example $1$ then $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges absolutely for z = $-1$ and diverges for z = $1$. But the statement above says we can go with $d$ as close to 0 as we want and we will still find a $z$ with $|z|=1$ for which $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges can somebody please show me the proof for all those questions I have?
Thank you!
sequences-and-series power-series
asked Jan 4 at 23:30
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
The convergenceradius is $1$ that means if $|z|= 1$ then the convergencebehaviour depends on $d$.
I have several questions about this issue:
1.Why is $sum_{n=0}^{infty}frac{1}{n^d}$ absolutely convergent for $d>1$ and $|z|=1$
2.Why is the implication true:
series absolutely divergent $Rightarrow$ series is divergent
3.Why is $sum_{n=0}^{infty}frac{1}{n^d}z^n$ divergent for $dleq 0$ and $|z|=1$
4.Why does the convergencebehaviour depends on $zin{ainmathbb{C}||a|=1}$ for $0<dleq 1$?
I know that if $d$ is for example $1$ then $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges absolutely for z = $-1$ and diverges for z = $1$. But the statement above says we can go with $d$ as close to 0 as we want and we will still find a $z$ with $|z|=1$ for which $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges can somebody please show me the proof for all those questions I have?
Thank you!
sequences-and-series power-series
asked Jan 4 at 23:30
RM777RM777
38312
$endgroup$
1
$begingroup$
Not from $n=0$.
$endgroup$
– Did
Jan 5 at 12:39
$begingroup$
@Did Question 1-3 were answered. Can you help me with my fourth question? $forall dinmathbb{R},0<dleq1$ there exist a $|z|=1,zneq 1$ such that $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converge conditionally
$endgroup$
– RM777
Jan 5 at 12:46
$begingroup$
Please replace "there exists $|z|=1$, $zne1$ such that..." by "for every $|z|=1$, $zne1$, one has..." What did you try to solve this?
$endgroup$
– Did
Jan 5 at 14:41
$begingroup$
Note that $$sum_{ngeq1}frac{z^n}{n^d}=mathrm{Li}_{d}(z)$$ where $mathrm{Li}$ is the poly-logarithm function
$endgroup$
– clathratus
Jan 7 at 19:08
add a comment |
$begingroup$
The convergenceradius is $1$ that means if $|z|= 1$ then the convergencebehaviour depends on $d$.
I have several questions about this issue:
1.Why is $sum_{n=0}^{infty}frac{1}{n^d}$ absolutely convergent for $d>1$ and $|z|=1$
2.Why is the implication true:
series absolutely divergent $Rightarrow$ series is divergent
3.Why is $sum_{n=0}^{infty}frac{1}{n^d}z^n$ divergent for $dleq 0$ and $|z|=1$
4.Why does the convergencebehaviour depends on $zin{ainmathbb{C}||a|=1}$ for $0<dleq 1$?
I know that if $d$ is for example $1$ then $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges absolutely for z = $-1$ and diverges for z = $1$. But the statement above says we can go with $d$ as close to 0 as we want and we will still find a $z$ with $|z|=1$ for which $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges can somebody please show me the proof for all those questions I have?
Thank you!
sequences-and-series power-series
asked Jan 4 at 23:30
RM777RM777
38312
$endgroup$
The convergenceradius is $1$ that means if $|z|= 1$ then the convergencebehaviour depends on $d$.
I have several questions about this issue:
1.Why is $sum_{n=0}^{infty}frac{1}{n^d}$ absolutely convergent for $d>1$ and $|z|=1$
2.Why is the implication true:
series absolutely divergent $Rightarrow$ series is divergent
3.Why is $sum_{n=0}^{infty}frac{1}{n^d}z^n$ divergent for $dleq 0$ and $|z|=1$
4.Why does the convergencebehaviour depends on $zin{ainmathbb{C}||a|=1}$ for $0<dleq 1$?
I know that if $d$ is for example $1$ then $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges absolutely for z = $-1$ and diverges for z = $1$. But the statement above says we can go with $d$ as close to 0 as we want and we will still find a $z$ with $|z|=1$ for which $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges can somebody please show me the proof for all those questions I have?
Thank you!
sequences-and-series power-series
sequences-and-series power-series
asked Jan 4 at 23:30
RM777RM777
38312
asked Jan 4 at 23:30
RM777RM777
38312
edited Jan 5 at 12:28
RM777
asked Jan 4 at 23:30
RM777RM777
38312
asked Jan 4 at 23:30
RM777RM777
38312
asked Jan 4 at 23:30
RM777RM777
38312
38312
1
$begingroup$
Not from $n=0$.
$endgroup$
– Did
Jan 5 at 12:39
$begingroup$
@Did Question 1-3 were answered. Can you help me with my fourth question? $forall dinmathbb{R},0<dleq1$ there exist a $|z|=1,zneq 1$ such that $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converge conditionally
$endgroup$
– RM777
Jan 5 at 12:46
$begingroup$
Please replace "there exists $|z|=1$, $zne1$ such that..." by "for every $|z|=1$, $zne1$, one has..." What did you try to solve this?
$endgroup$
– Did
Jan 5 at 14:41
$begingroup$
Note that $$sum_{ngeq1}frac{z^n}{n^d}=mathrm{Li}_{d}(z)$$ where $mathrm{Li}$ is the poly-logarithm function
$endgroup$
– clathratus
Jan 7 at 19:08
add a comment |
1
$begingroup$
Not from $n=0$.
$endgroup$
– Did
Jan 5 at 12:39
$begingroup$
@Did Question 1-3 were answered. Can you help me with my fourth question? $forall dinmathbb{R},0<dleq1$ there exist a $|z|=1,zneq 1$ such that $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converge conditionally
$endgroup$
– RM777
Jan 5 at 12:46
$begingroup$
Please replace "there exists $|z|=1$, $zne1$ such that..." by "for every $|z|=1$, $zne1$, one has..." What did you try to solve this?
$endgroup$
– Did
Jan 5 at 14:41
$begingroup$
Note that $$sum_{ngeq1}frac{z^n}{n^d}=mathrm{Li}_{d}(z)$$ where $mathrm{Li}$ is the poly-logarithm function
$endgroup$
– clathratus
Jan 7 at 19:08
1
1
$begingroup$
Not from $n=0$.
$endgroup$
– Did
Jan 5 at 12:39
$begingroup$
Not from $n=0$.
$endgroup$
– Did
Jan 5 at 12:39
$begingroup$
@Did Question 1-3 were answered. Can you help me with my fourth question? $forall dinmathbb{R},0<dleq1$ there exist a $|z|=1,zneq 1$ such that $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converge conditionally
$endgroup$
– RM777
Jan 5 at 12:46
$begingroup$
@Did Question 1-3 were answered. Can you help me with my fourth question? $forall dinmathbb{R},0<dleq1$ there exist a $|z|=1,zneq 1$ such that $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converge conditionally
$endgroup$
– RM777
Jan 5 at 12:46
$begingroup$
Please replace "there exists $|z|=1$, $zne1$ such that..." by "for every $|z|=1$, $zne1$, one has..." What did you try to solve this?
$endgroup$
– Did
Jan 5 at 14:41
$begingroup$
Please replace "there exists $|z|=1$, $zne1$ such that..." by "for every $|z|=1$, $zne1$, one has..." What did you try to solve this?
$endgroup$
– Did
Jan 5 at 14:41
$begingroup$
Note that $$sum_{ngeq1}frac{z^n}{n^d}=mathrm{Li}_{d}(z)$$ where $mathrm{Li}$ is the poly-logarithm function
$endgroup$
– clathratus
Jan 7 at 19:08
$begingroup$
Note that $$sum_{ngeq1}frac{z^n}{n^d}=mathrm{Li}_{d}(z)$$ where $mathrm{Li}$ is the poly-logarithm function
$endgroup$
– clathratus
Jan 7 at 19:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
- Apply integral test. (Thanks to T. Bongers for pointing out an error I made earlier).
- False. $sum frac {(-1)^{n}} n$ is absolutely divergent but not divergent.
- The general term does not tend to $0$ so the series is divergent.
- For $0<d<1$ the series diverges absolutely whenever $|z|=1$. [Compare with $sum frac 1 n$]. The series converges for $z=-1$ and diverges for $z=1$. For $z=-1$ use Alternating Series Test.
answered Jan 4 at 23:36
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
$begingroup$
Ratio test fails for (1). Integral test works.
$endgroup$
– T. Bongers
Jan 4 at 23:43
$begingroup$
Sorry in 4 I actually wanted to say $0<dleq1$. My question was if I would take a smallpositive number in this intervall $epsilon$ how can I proof existence of a $zinmathbb{C}:|z|=1$ and $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges absolutely.
$endgroup$
– RM777
Jan 4 at 23:45
$begingroup$
@RM777 You cannot. If $d$ is a small positive number the series does not converge absolutely for any $z$ with $|z|=1$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:48
$begingroup$
I am unfammiliar with the integral test, can I apply also the root criteria?
$endgroup$
– RM777
Jan 4 at 23:49
$begingroup$
@RM777 No. Root test and ratio test both fail in this case.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:50
|
show 4 more comments
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%2f3062236%2fquestions-about-the-powerseries-sum-n-0-infty-frac1ndzn-d-in-mathbb%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
$begingroup$
- Apply integral test. (Thanks to T. Bongers for pointing out an error I made earlier).
- False. $sum frac {(-1)^{n}} n$ is absolutely divergent but not divergent.
- The general term does not tend to $0$ so the series is divergent.
- For $0<d<1$ the series diverges absolutely whenever $|z|=1$. [Compare with $sum frac 1 n$]. The series converges for $z=-1$ and diverges for $z=1$. For $z=-1$ use Alternating Series Test.
answered Jan 4 at 23:36
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
$begingroup$
Ratio test fails for (1). Integral test works.
$endgroup$
– T. Bongers
Jan 4 at 23:43
$begingroup$
Sorry in 4 I actually wanted to say $0<dleq1$. My question was if I would take a smallpositive number in this intervall $epsilon$ how can I proof existence of a $zinmathbb{C}:|z|=1$ and $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges absolutely.
$endgroup$
– RM777
Jan 4 at 23:45
$begingroup$
@RM777 You cannot. If $d$ is a small positive number the series does not converge absolutely for any $z$ with $|z|=1$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:48
$begingroup$
I am unfammiliar with the integral test, can I apply also the root criteria?
$endgroup$
– RM777
Jan 4 at 23:49
$begingroup$
@RM777 No. Root test and ratio test both fail in this case.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:50
|
show 4 more comments
$begingroup$
- Apply integral test. (Thanks to T. Bongers for pointing out an error I made earlier).
- False. $sum frac {(-1)^{n}} n$ is absolutely divergent but not divergent.
- The general term does not tend to $0$ so the series is divergent.
- For $0<d<1$ the series diverges absolutely whenever $|z|=1$. [Compare with $sum frac 1 n$]. The series converges for $z=-1$ and diverges for $z=1$. For $z=-1$ use Alternating Series Test.
answered Jan 4 at 23:36
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
$begingroup$
Ratio test fails for (1). Integral test works.
$endgroup$
– T. Bongers
Jan 4 at 23:43
$begingroup$
Sorry in 4 I actually wanted to say $0<dleq1$. My question was if I would take a smallpositive number in this intervall $epsilon$ how can I proof existence of a $zinmathbb{C}:|z|=1$ and $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges absolutely.
$endgroup$
– RM777
Jan 4 at 23:45
$begingroup$
@RM777 You cannot. If $d$ is a small positive number the series does not converge absolutely for any $z$ with $|z|=1$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:48
$begingroup$
I am unfammiliar with the integral test, can I apply also the root criteria?
$endgroup$
– RM777
Jan 4 at 23:49
$begingroup$
@RM777 No. Root test and ratio test both fail in this case.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:50
|
show 4 more comments
$begingroup$
- Apply integral test. (Thanks to T. Bongers for pointing out an error I made earlier).
- False. $sum frac {(-1)^{n}} n$ is absolutely divergent but not divergent.
- The general term does not tend to $0$ so the series is divergent.
- For $0<d<1$ the series diverges absolutely whenever $|z|=1$. [Compare with $sum frac 1 n$]. The series converges for $z=-1$ and diverges for $z=1$. For $z=-1$ use Alternating Series Test.
answered Jan 4 at 23:36
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
- Apply integral test. (Thanks to T. Bongers for pointing out an error I made earlier).
- False. $sum frac {(-1)^{n}} n$ is absolutely divergent but not divergent.
- The general term does not tend to $0$ so the series is divergent.
- For $0<d<1$ the series diverges absolutely whenever $|z|=1$. [Compare with $sum frac 1 n$]. The series converges for $z=-1$ and diverges for $z=1$. For $z=-1$ use Alternating Series Test.
answered Jan 4 at 23:36
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
edited Jan 4 at 23:56
answered Jan 4 at 23:36
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 4 at 23:36
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 4 at 23:36
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
72.6k53170
$begingroup$
Ratio test fails for (1). Integral test works.
$endgroup$
– T. Bongers
Jan 4 at 23:43
$begingroup$
Sorry in 4 I actually wanted to say $0<dleq1$. My question was if I would take a smallpositive number in this intervall $epsilon$ how can I proof existence of a $zinmathbb{C}:|z|=1$ and $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges absolutely.
$endgroup$
– RM777
Jan 4 at 23:45
$begingroup$
@RM777 You cannot. If $d$ is a small positive number the series does not converge absolutely for any $z$ with $|z|=1$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:48
$begingroup$
I am unfammiliar with the integral test, can I apply also the root criteria?
$endgroup$
– RM777
Jan 4 at 23:49
$begingroup$
@RM777 No. Root test and ratio test both fail in this case.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:50
|
show 4 more comments
$begingroup$
Ratio test fails for (1). Integral test works.
$endgroup$
– T. Bongers
Jan 4 at 23:43
$begingroup$
Sorry in 4 I actually wanted to say $0<dleq1$. My question was if I would take a smallpositive number in this intervall $epsilon$ how can I proof existence of a $zinmathbb{C}:|z|=1$ and $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges absolutely.
$endgroup$
– RM777
Jan 4 at 23:45
$begingroup$
@RM777 You cannot. If $d$ is a small positive number the series does not converge absolutely for any $z$ with $|z|=1$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:48
$begingroup$
I am unfammiliar with the integral test, can I apply also the root criteria?
$endgroup$
– RM777
Jan 4 at 23:49
$begingroup$
@RM777 No. Root test and ratio test both fail in this case.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:50
$begingroup$
Ratio test fails for (1). Integral test works.
$endgroup$
– T. Bongers
Jan 4 at 23:43
$begingroup$
Ratio test fails for (1). Integral test works.
$endgroup$
– T. Bongers
Jan 4 at 23:43
$begingroup$
Sorry in 4 I actually wanted to say $0<dleq1$. My question was if I would take a smallpositive number in this intervall $epsilon$ how can I proof existence of a $zinmathbb{C}:|z|=1$ and $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges absolutely.
$endgroup$
– RM777
Jan 4 at 23:45
$begingroup$
Sorry in 4 I actually wanted to say $0<dleq1$. My question was if I would take a smallpositive number in this intervall $epsilon$ how can I proof existence of a $zinmathbb{C}:|z|=1$ and $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converges absolutely.
$endgroup$
– RM777
Jan 4 at 23:45
$begingroup$
@RM777 You cannot. If $d$ is a small positive number the series does not converge absolutely for any $z$ with $|z|=1$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:48
$begingroup$
@RM777 You cannot. If $d$ is a small positive number the series does not converge absolutely for any $z$ with $|z|=1$.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:48
$begingroup$
I am unfammiliar with the integral test, can I apply also the root criteria?
$endgroup$
– RM777
Jan 4 at 23:49
$begingroup$
I am unfammiliar with the integral test, can I apply also the root criteria?
$endgroup$
– RM777
Jan 4 at 23:49
$begingroup$
@RM777 No. Root test and ratio test both fail in this case.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:50
$begingroup$
@RM777 No. Root test and ratio test both fail in this case.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 23:50
|
show 4 more comments
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%2f3062236%2fquestions-about-the-powerseries-sum-n-0-infty-frac1ndzn-d-in-mathbb%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
$begingroup$
Not from $n=0$.
$endgroup$
– Did
Jan 5 at 12:39
$begingroup$
@Did Question 1-3 were answered. Can you help me with my fourth question? $forall dinmathbb{R},0<dleq1$ there exist a $|z|=1,zneq 1$ such that $sum_{n=0}^{infty}frac{1}{n^d}z^n$ converge conditionally
$endgroup$
– RM777
Jan 5 at 12:46
$begingroup$
Please replace "there exists $|z|=1$, $zne1$ such that..." by "for every $|z|=1$, $zne1$, one has..." What did you try to solve this?
$endgroup$
– Did
Jan 5 at 14:41
$begingroup$
Note that $$sum_{ngeq1}frac{z^n}{n^d}=mathrm{Li}_{d}(z)$$ where $mathrm{Li}$ is the poly-logarithm function
$endgroup$
– clathratus
Jan 7 at 19:08