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 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


















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

3,816623




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


















  • $begingroup$
    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











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
    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
    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

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




















  • $begingroup$
    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




















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

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more