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







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






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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














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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




















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