Convergence of a complex series to a function with double poles on positive integers
$begingroup$
Prove that the series $sum_{z=1}^{infty} frac {1}{(z-n)^2}$ converges on the complex plane minus the positive integers to an analytic function with a double pole at each positive integer.
Now by limit comparison test with $frac {1}{n^2}$ we know that the series is absolutely convergent except at positive integers where it is not defined. More specifically, I am using that $$lim_{n to infty} frac{|frac {1}{(z-n)^2} |}{|frac{1}{n^2}|} = 1$$ I am not sure how to proceed from here.
complex-analysis
$endgroup$
add a comment |
$begingroup$
Prove that the series $sum_{z=1}^{infty} frac {1}{(z-n)^2}$ converges on the complex plane minus the positive integers to an analytic function with a double pole at each positive integer.
Now by limit comparison test with $frac {1}{n^2}$ we know that the series is absolutely convergent except at positive integers where it is not defined. More specifically, I am using that $$lim_{n to infty} frac{|frac {1}{(z-n)^2} |}{|frac{1}{n^2}|} = 1$$ I am not sure how to proceed from here.
complex-analysis
$endgroup$
$begingroup$
$sum_{n=1}^N frac{1}{(z-n)^2}$ is analytic for each $N$ and converges locally uniformly to $sum_{n=1}^infty frac{1}{(z-n)^2}$
$endgroup$
– mathworker21
Jan 2 at 11:36
add a comment |
$begingroup$
Prove that the series $sum_{z=1}^{infty} frac {1}{(z-n)^2}$ converges on the complex plane minus the positive integers to an analytic function with a double pole at each positive integer.
Now by limit comparison test with $frac {1}{n^2}$ we know that the series is absolutely convergent except at positive integers where it is not defined. More specifically, I am using that $$lim_{n to infty} frac{|frac {1}{(z-n)^2} |}{|frac{1}{n^2}|} = 1$$ I am not sure how to proceed from here.
complex-analysis
$endgroup$
Prove that the series $sum_{z=1}^{infty} frac {1}{(z-n)^2}$ converges on the complex plane minus the positive integers to an analytic function with a double pole at each positive integer.
Now by limit comparison test with $frac {1}{n^2}$ we know that the series is absolutely convergent except at positive integers where it is not defined. More specifically, I am using that $$lim_{n to infty} frac{|frac {1}{(z-n)^2} |}{|frac{1}{n^2}|} = 1$$ I am not sure how to proceed from here.
complex-analysis
complex-analysis
asked Jan 2 at 11:32
David WarrenDavid Warren
586314
586314
$begingroup$
$sum_{n=1}^N frac{1}{(z-n)^2}$ is analytic for each $N$ and converges locally uniformly to $sum_{n=1}^infty frac{1}{(z-n)^2}$
$endgroup$
– mathworker21
Jan 2 at 11:36
add a comment |
$begingroup$
$sum_{n=1}^N frac{1}{(z-n)^2}$ is analytic for each $N$ and converges locally uniformly to $sum_{n=1}^infty frac{1}{(z-n)^2}$
$endgroup$
– mathworker21
Jan 2 at 11:36
$begingroup$
$sum_{n=1}^N frac{1}{(z-n)^2}$ is analytic for each $N$ and converges locally uniformly to $sum_{n=1}^infty frac{1}{(z-n)^2}$
$endgroup$
– mathworker21
Jan 2 at 11:36
$begingroup$
$sum_{n=1}^N frac{1}{(z-n)^2}$ is analytic for each $N$ and converges locally uniformly to $sum_{n=1}^infty frac{1}{(z-n)^2}$
$endgroup$
– mathworker21
Jan 2 at 11:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hints:
1) Prove that $sumlimits_{n=1}^infty dfrac{1}{(z-n)^2}$ is well defined, ie, that defines a function $f(z)$ for all $zinmathbb{C}-mathbb{N}$
2)Note that, for each $minmathbb{N}$, $$sumlimits_{n=1}^infty dfrac{1}{(z-n)^2}=dfrac{1}{(z-m)^2}+underbrace{sumlimits_{n=1,nneq m}^infty dfrac{1}{(z-n)^2}}_{text{holomorphic in }V_m}$$ where $V_m$ is a neighborhood of $m$ (so that m is the only singularity there).
3) Remember the definition of pole
$endgroup$
$begingroup$
To prove that it is well-defined, I need to show that the sum converges. I'm not sure how that follows
$endgroup$
– David Warren
Jan 4 at 5:46
add a comment |
$begingroup$
In addition to what Martin Vacas Vignolo has suggested here is the argument needed to prove that series converges to an analytic function in $mathbb C setminus mathbb N$: if $K subset mathbb C setminus mathbb N$ is compact then $|(z-n)^{2}|geq (n -|z|)^{2}$. Use the fact that $K$ is bounded and ignore the first few terms to conclude that the series converges uniformly on $K$. This proves that $sum frac 1 {(z-n)^{2}}$ is analytic in $mathbb C setminus mathbb N$.
$endgroup$
$begingroup$
which theorem are you invoking? how do you know it converges uniformly?
$endgroup$
– David Warren
Jan 3 at 23:42
$begingroup$
@DavidWarren Use M-test. $sum_{n=[a]+1}^{infty} frac 1 {(n-a)^{2}} <infty$.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:44
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%2f3059370%2fconvergence-of-a-complex-series-to-a-function-with-double-poles-on-positive-inte%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
$begingroup$
Hints:
1) Prove that $sumlimits_{n=1}^infty dfrac{1}{(z-n)^2}$ is well defined, ie, that defines a function $f(z)$ for all $zinmathbb{C}-mathbb{N}$
2)Note that, for each $minmathbb{N}$, $$sumlimits_{n=1}^infty dfrac{1}{(z-n)^2}=dfrac{1}{(z-m)^2}+underbrace{sumlimits_{n=1,nneq m}^infty dfrac{1}{(z-n)^2}}_{text{holomorphic in }V_m}$$ where $V_m$ is a neighborhood of $m$ (so that m is the only singularity there).
3) Remember the definition of pole
$endgroup$
$begingroup$
To prove that it is well-defined, I need to show that the sum converges. I'm not sure how that follows
$endgroup$
– David Warren
Jan 4 at 5:46
add a comment |
$begingroup$
Hints:
1) Prove that $sumlimits_{n=1}^infty dfrac{1}{(z-n)^2}$ is well defined, ie, that defines a function $f(z)$ for all $zinmathbb{C}-mathbb{N}$
2)Note that, for each $minmathbb{N}$, $$sumlimits_{n=1}^infty dfrac{1}{(z-n)^2}=dfrac{1}{(z-m)^2}+underbrace{sumlimits_{n=1,nneq m}^infty dfrac{1}{(z-n)^2}}_{text{holomorphic in }V_m}$$ where $V_m$ is a neighborhood of $m$ (so that m is the only singularity there).
3) Remember the definition of pole
$endgroup$
$begingroup$
To prove that it is well-defined, I need to show that the sum converges. I'm not sure how that follows
$endgroup$
– David Warren
Jan 4 at 5:46
add a comment |
$begingroup$
Hints:
1) Prove that $sumlimits_{n=1}^infty dfrac{1}{(z-n)^2}$ is well defined, ie, that defines a function $f(z)$ for all $zinmathbb{C}-mathbb{N}$
2)Note that, for each $minmathbb{N}$, $$sumlimits_{n=1}^infty dfrac{1}{(z-n)^2}=dfrac{1}{(z-m)^2}+underbrace{sumlimits_{n=1,nneq m}^infty dfrac{1}{(z-n)^2}}_{text{holomorphic in }V_m}$$ where $V_m$ is a neighborhood of $m$ (so that m is the only singularity there).
3) Remember the definition of pole
$endgroup$
Hints:
1) Prove that $sumlimits_{n=1}^infty dfrac{1}{(z-n)^2}$ is well defined, ie, that defines a function $f(z)$ for all $zinmathbb{C}-mathbb{N}$
2)Note that, for each $minmathbb{N}$, $$sumlimits_{n=1}^infty dfrac{1}{(z-n)^2}=dfrac{1}{(z-m)^2}+underbrace{sumlimits_{n=1,nneq m}^infty dfrac{1}{(z-n)^2}}_{text{holomorphic in }V_m}$$ where $V_m$ is a neighborhood of $m$ (so that m is the only singularity there).
3) Remember the definition of pole
answered Jan 2 at 11:46
Martín Vacas VignoloMartín Vacas Vignolo
3,816623
3,816623
$begingroup$
To prove that it is well-defined, I need to show that the sum converges. I'm not sure how that follows
$endgroup$
– David Warren
Jan 4 at 5:46
add a comment |
$begingroup$
To prove that it is well-defined, I need to show that the sum converges. I'm not sure how that follows
$endgroup$
– David Warren
Jan 4 at 5:46
$begingroup$
To prove that it is well-defined, I need to show that the sum converges. I'm not sure how that follows
$endgroup$
– David Warren
Jan 4 at 5:46
$begingroup$
To prove that it is well-defined, I need to show that the sum converges. I'm not sure how that follows
$endgroup$
– David Warren
Jan 4 at 5:46
add a comment |
$begingroup$
In addition to what Martin Vacas Vignolo has suggested here is the argument needed to prove that series converges to an analytic function in $mathbb C setminus mathbb N$: if $K subset mathbb C setminus mathbb N$ is compact then $|(z-n)^{2}|geq (n -|z|)^{2}$. Use the fact that $K$ is bounded and ignore the first few terms to conclude that the series converges uniformly on $K$. This proves that $sum frac 1 {(z-n)^{2}}$ is analytic in $mathbb C setminus mathbb N$.
$endgroup$
$begingroup$
which theorem are you invoking? how do you know it converges uniformly?
$endgroup$
– David Warren
Jan 3 at 23:42
$begingroup$
@DavidWarren Use M-test. $sum_{n=[a]+1}^{infty} frac 1 {(n-a)^{2}} <infty$.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:44
add a comment |
$begingroup$
In addition to what Martin Vacas Vignolo has suggested here is the argument needed to prove that series converges to an analytic function in $mathbb C setminus mathbb N$: if $K subset mathbb C setminus mathbb N$ is compact then $|(z-n)^{2}|geq (n -|z|)^{2}$. Use the fact that $K$ is bounded and ignore the first few terms to conclude that the series converges uniformly on $K$. This proves that $sum frac 1 {(z-n)^{2}}$ is analytic in $mathbb C setminus mathbb N$.
$endgroup$
$begingroup$
which theorem are you invoking? how do you know it converges uniformly?
$endgroup$
– David Warren
Jan 3 at 23:42
$begingroup$
@DavidWarren Use M-test. $sum_{n=[a]+1}^{infty} frac 1 {(n-a)^{2}} <infty$.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:44
add a comment |
$begingroup$
In addition to what Martin Vacas Vignolo has suggested here is the argument needed to prove that series converges to an analytic function in $mathbb C setminus mathbb N$: if $K subset mathbb C setminus mathbb N$ is compact then $|(z-n)^{2}|geq (n -|z|)^{2}$. Use the fact that $K$ is bounded and ignore the first few terms to conclude that the series converges uniformly on $K$. This proves that $sum frac 1 {(z-n)^{2}}$ is analytic in $mathbb C setminus mathbb N$.
$endgroup$
In addition to what Martin Vacas Vignolo has suggested here is the argument needed to prove that series converges to an analytic function in $mathbb C setminus mathbb N$: if $K subset mathbb C setminus mathbb N$ is compact then $|(z-n)^{2}|geq (n -|z|)^{2}$. Use the fact that $K$ is bounded and ignore the first few terms to conclude that the series converges uniformly on $K$. This proves that $sum frac 1 {(z-n)^{2}}$ is analytic in $mathbb C setminus mathbb N$.
answered Jan 2 at 11:55
Kavi Rama MurthyKavi Rama Murthy
63.7k42464
63.7k42464
$begingroup$
which theorem are you invoking? how do you know it converges uniformly?
$endgroup$
– David Warren
Jan 3 at 23:42
$begingroup$
@DavidWarren Use M-test. $sum_{n=[a]+1}^{infty} frac 1 {(n-a)^{2}} <infty$.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:44
add a comment |
$begingroup$
which theorem are you invoking? how do you know it converges uniformly?
$endgroup$
– David Warren
Jan 3 at 23:42
$begingroup$
@DavidWarren Use M-test. $sum_{n=[a]+1}^{infty} frac 1 {(n-a)^{2}} <infty$.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:44
$begingroup$
which theorem are you invoking? how do you know it converges uniformly?
$endgroup$
– David Warren
Jan 3 at 23:42
$begingroup$
which theorem are you invoking? how do you know it converges uniformly?
$endgroup$
– David Warren
Jan 3 at 23:42
$begingroup$
@DavidWarren Use M-test. $sum_{n=[a]+1}^{infty} frac 1 {(n-a)^{2}} <infty$.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:44
$begingroup$
@DavidWarren Use M-test. $sum_{n=[a]+1}^{infty} frac 1 {(n-a)^{2}} <infty$.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:44
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%2f3059370%2fconvergence-of-a-complex-series-to-a-function-with-double-poles-on-positive-inte%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
$begingroup$
$sum_{n=1}^N frac{1}{(z-n)^2}$ is analytic for each $N$ and converges locally uniformly to $sum_{n=1}^infty frac{1}{(z-n)^2}$
$endgroup$
– mathworker21
Jan 2 at 11:36