I was looking into a previous exam from 2011 of a course I am taking of Complex Analysis, and they askWhich of the following series converge to a rational function in some domain? ∑ k = 0 ∞ 1 k ! + k 2 + k z k 2 + 2 k ∑ k = 0 ∞ 2 k ( 1 + z 2 ) k ∑ k = 0 ∞ 1 k ! ( z − k ) k I had no problem with the second and third ones. The second one converges to 1 1 − 2 1 + z 2 given that | 2 1 + z 2 | &lt; 1, and the third one has an infinite number of singularities so it can't be a rational function.I don't know how to verify that in the first series. Is there any general method to see if some power series converge to a rational function, or in this particular case, a way to see if this series does?Edit: Now I am not that convinced about my argument for the third series (Since the series might only converge in a bounded domain from which the number of singularities would be finite). Is there anything wrong with it or any way to formalize it further?

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I was looking into a previous exam from 2011 of a course I am taking of Complex Analysis, and they askWhich of the following series converge to a rational function in some domain?      ∑          k      =      0        ∞        1          k      !      +              k        2            +      k            z                  k        2            +      2      k              ∑          k      =      0        ∞              2      k              (      1      +              z        2                    )        k                    ∑          k      =      0        ∞        1          k      !      (      z      −      k              )        k            I had no problem with the second and third ones. The second one converges to       1          1      −              2                  1          +                      z            2                               given that       |          2              1        +                  z          2                      |    &amp;lt;  1, and the third one has an infinite number of singularities so it can&#039;t be a rational function.I don&#039;t know how to verify that in the first series. Is there any general method to see if some power series converge to a rational function, or in this particular case, a way to see if this series does?Edit: Now I am not that convinced about my argument for the third series (Since the series might only converge in a bounded domain from which the number of singularities would be finite). Is there anything wrong with it or any way to formalize it further?

esperoanow · Accepted Answer

The first series converges uniformly to some   f in             D        ¯    , and diverges for all other   z. Thus   f is analytic in   D. Suppose   f  =  R in some domain   U, where   R is a rational function. Then   U  ⊂      D    .. Let   P be the set of poles of   R  .. Then   f  ,  R are both analytic in       D    ∖  P  ,, which is also a domain. By the identity principle,   f  =  R in       D    ∖  P  . Since   f has no singularities in   D  (  0  ,  1  )  , neither does   R  .. Therefore   f  =  R in       D    .. Now   f extends continuously to             D        ¯    , hence so does   R  . Therefore   R has no poles in             D        ¯    . It follows that   R is analytic in some   D  (  0  ,  r  )  ,  r  &amp;gt;  1. Thus   f has an analytic extension to   D  (  0  ,  r  )  . Therefore the power series defining   f converges in   D  (  0  ,  r  )  , contradiction.Added later For the third series: Let   Ω  =      C    ∖  {  0  ,  1  ,  2  ,  …  }  . Note that   Ω is an open connected set. The given series converges uniformly on compact subsets of   Ω to an analytic function   f, and   f has a pole at each point of   {  0  ,  1  ,  2  ,  …  }  . (Please ask if you have questions on this.) Let   R be a rational function, and let   P be the set of poles of   R. Then both   f  ,  R are analytic on   Ω  ∖  P  , which is also an open connected set. Suppose   f  =  R on an open subset of   Ω  ∖  P  . Then by the identity principle,   f  =  R everywhere in   Ω  ∖  P  . It follows that   R has a pole at each point of   {  0  ,  1  ,  2  ,  …  }  . This is a contradiction, since   R has at most a finite number of poles.

I was looking into a previous exam from 2011 of a course I am taking of Complex Analysis, and they a

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