Analytic properties of a series











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I was wondering about whether it is possible that the analytic properties of a series are different from the ones of the partial sums. For instance suppose we have the series
$$S=sum_{n=0}^{infty}frac{1}{z-c_n}$$
where $c_n$ are positive real numbers.



Clearly the partial sums
$$S_N=sum_{n=0}^{N}frac{1}{z-c_n}$$
are analytic functions with a collection of simple poles on the real line.



Is it possible that the series develops branch cuts or something it isn't possible to infer from the partial sum? An anser would be welcome even if it isn't related to the particular example I produced.










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    I was wondering about whether it is possible that the analytic properties of a series are different from the ones of the partial sums. For instance suppose we have the series
    $$S=sum_{n=0}^{infty}frac{1}{z-c_n}$$
    where $c_n$ are positive real numbers.



    Clearly the partial sums
    $$S_N=sum_{n=0}^{N}frac{1}{z-c_n}$$
    are analytic functions with a collection of simple poles on the real line.



    Is it possible that the series develops branch cuts or something it isn't possible to infer from the partial sum? An anser would be welcome even if it isn't related to the particular example I produced.










    share|cite|improve this question
























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      I was wondering about whether it is possible that the analytic properties of a series are different from the ones of the partial sums. For instance suppose we have the series
      $$S=sum_{n=0}^{infty}frac{1}{z-c_n}$$
      where $c_n$ are positive real numbers.



      Clearly the partial sums
      $$S_N=sum_{n=0}^{N}frac{1}{z-c_n}$$
      are analytic functions with a collection of simple poles on the real line.



      Is it possible that the series develops branch cuts or something it isn't possible to infer from the partial sum? An anser would be welcome even if it isn't related to the particular example I produced.










      share|cite|improve this question













      I was wondering about whether it is possible that the analytic properties of a series are different from the ones of the partial sums. For instance suppose we have the series
      $$S=sum_{n=0}^{infty}frac{1}{z-c_n}$$
      where $c_n$ are positive real numbers.



      Clearly the partial sums
      $$S_N=sum_{n=0}^{N}frac{1}{z-c_n}$$
      are analytic functions with a collection of simple poles on the real line.



      Is it possible that the series develops branch cuts or something it isn't possible to infer from the partial sum? An anser would be welcome even if it isn't related to the particular example I produced.







      sequences-and-series analyticity






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      asked Dec 3 at 18:29









      Fra

      24615




      24615






















          2 Answers
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          Another is



          $sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
          =dfrac{pi}{sin{pi z}}
          $
          .






          share|cite|improve this answer




























            up vote
            0
            down vote













            The doubt came and went rather fast. The answer seems to be yes, the series can have analytic properties different from the partial sum. Since I don't have the means to prove it in general, I thought an example might just do the trick to show that it is possible.



            Consider the case in which $c_n= a n^2 +b$ where $a,bin mathbb{R}^+$, then
            $$S(z)=sum_{n=0}^{infty}frac{1}{a n^2+b-z}=frac{pi sqrt{z-b} cot left(frac{pi sqrt{z-b}}{sqrt{a}}right)-sqrt{a}}{2 sqrt{a} (b-z)}$$
            which clearly shows that $S(z)$ has a branch cut, while the partial sum does not.






            share|cite|improve this answer





















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote













              Another is



              $sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
              =dfrac{pi}{sin{pi z}}
              $
              .






              share|cite|improve this answer

























                up vote
                1
                down vote













                Another is



                $sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
                =dfrac{pi}{sin{pi z}}
                $
                .






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Another is



                  $sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
                  =dfrac{pi}{sin{pi z}}
                  $
                  .






                  share|cite|improve this answer












                  Another is



                  $sumlimits_{n=-infty}^{infty} dfrac{(-1)^n}{z-n}
                  =dfrac{pi}{sin{pi z}}
                  $
                  .







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 3 at 22:12









                  marty cohen

                  71.7k546124




                  71.7k546124






















                      up vote
                      0
                      down vote













                      The doubt came and went rather fast. The answer seems to be yes, the series can have analytic properties different from the partial sum. Since I don't have the means to prove it in general, I thought an example might just do the trick to show that it is possible.



                      Consider the case in which $c_n= a n^2 +b$ where $a,bin mathbb{R}^+$, then
                      $$S(z)=sum_{n=0}^{infty}frac{1}{a n^2+b-z}=frac{pi sqrt{z-b} cot left(frac{pi sqrt{z-b}}{sqrt{a}}right)-sqrt{a}}{2 sqrt{a} (b-z)}$$
                      which clearly shows that $S(z)$ has a branch cut, while the partial sum does not.






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        The doubt came and went rather fast. The answer seems to be yes, the series can have analytic properties different from the partial sum. Since I don't have the means to prove it in general, I thought an example might just do the trick to show that it is possible.



                        Consider the case in which $c_n= a n^2 +b$ where $a,bin mathbb{R}^+$, then
                        $$S(z)=sum_{n=0}^{infty}frac{1}{a n^2+b-z}=frac{pi sqrt{z-b} cot left(frac{pi sqrt{z-b}}{sqrt{a}}right)-sqrt{a}}{2 sqrt{a} (b-z)}$$
                        which clearly shows that $S(z)$ has a branch cut, while the partial sum does not.






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          The doubt came and went rather fast. The answer seems to be yes, the series can have analytic properties different from the partial sum. Since I don't have the means to prove it in general, I thought an example might just do the trick to show that it is possible.



                          Consider the case in which $c_n= a n^2 +b$ where $a,bin mathbb{R}^+$, then
                          $$S(z)=sum_{n=0}^{infty}frac{1}{a n^2+b-z}=frac{pi sqrt{z-b} cot left(frac{pi sqrt{z-b}}{sqrt{a}}right)-sqrt{a}}{2 sqrt{a} (b-z)}$$
                          which clearly shows that $S(z)$ has a branch cut, while the partial sum does not.






                          share|cite|improve this answer












                          The doubt came and went rather fast. The answer seems to be yes, the series can have analytic properties different from the partial sum. Since I don't have the means to prove it in general, I thought an example might just do the trick to show that it is possible.



                          Consider the case in which $c_n= a n^2 +b$ where $a,bin mathbb{R}^+$, then
                          $$S(z)=sum_{n=0}^{infty}frac{1}{a n^2+b-z}=frac{pi sqrt{z-b} cot left(frac{pi sqrt{z-b}}{sqrt{a}}right)-sqrt{a}}{2 sqrt{a} (b-z)}$$
                          which clearly shows that $S(z)$ has a branch cut, while the partial sum does not.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 3 at 20:50









                          Fra

                          24615




                          24615






























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