Convergence of a complex series to a function with double poles on positive integers












0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    $sum_{n=1}^N frac{1}{(z-n)^2}$ 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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    $sum_{n=1}^N frac{1}{(z-n)^2}$ 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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 2 at 11:32









David Warren KatzDavid Warren Katz

598315




598315












  • $begingroup$
    $sum_{n=1}^N frac{1}{(z-n)^2}$ 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




$begingroup$
$sum_{n=1}^N frac{1}{(z-n)^2}$ 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










2 Answers
2






active

oldest

votes


















2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 Katz
    Jan 4 at 5:46



















1












$begingroup$

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 $mathbb C setminus mathbb N$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    which theorem are you invoking? how do you know it converges uniformly?
    $endgroup$
    – David Warren Katz
    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














Your Answer








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%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









2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 Katz
    Jan 4 at 5:46
















2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 Katz
    Jan 4 at 5:46














2












2








2





$begingroup$

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






share|cite|improve this answer









$endgroup$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 11:46









Martín Vacas VignoloMartín Vacas Vignolo

3,796623




3,796623












  • $begingroup$
    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 Katz
    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 Katz
    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 Katz
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 Katz
Jan 4 at 5:46











1












$begingroup$

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 $mathbb C setminus mathbb N$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    which theorem are you invoking? how do you know it converges uniformly?
    $endgroup$
    – David Warren Katz
    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


















1












$begingroup$

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 $mathbb C setminus mathbb N$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    which theorem are you invoking? how do you know it converges uniformly?
    $endgroup$
    – David Warren Katz
    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
















1












1








1





$begingroup$

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 $mathbb C setminus mathbb N$.






share|cite|improve this answer









$endgroup$



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 $mathbb C setminus mathbb N$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 11:55









Kavi Rama MurthyKavi Rama Murthy

76.9k53471




76.9k53471












  • $begingroup$
    which theorem are you invoking? how do you know it converges uniformly?
    $endgroup$
    – David Warren Katz
    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 Katz
    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 Katz
Jan 3 at 23:42




$begingroup$
which theorem are you invoking? how do you know it converges uniformly?
$endgroup$
– David Warren Katz
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




















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%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





















































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

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more