From: stong@math.rice.edu (Richard Stong)
Newsgroups: sci.math.research
Subject: Taylor Series of  e^{z/(z-1)}
Date: 21 Mar 1997 15:50:15 GMT

Robert Dawson asked whether the Taylor series for e^{z/(z-1)} converges at
z=1.  In reply to this Robert Israel noted that doing some checking on MAPLE 
the nth Taylor coefficient is well approximated by

	a_n= n^{-3/4} (0.655 cos(2 sqrt(n) - 0.659 sin(2 sqrt(n)).

Further Israel remarked that if this property continues past the range where
he checked it then convergence occurs.

  This is relatively straightforward to check using the techniques of DeBruijn
Asymptotic Methods in Analysis, or Bender and Orszag, Advanced Mathematical
Methods for Scientists and Engineers.  The nth Taylor coefficient can be 
written as

	a_n= 1/(2 pi i) Integral  e^{z/(z-1)} z^{-(n+1)} dz

where the integral as usual is taken around any counterclockwise contour 
containing 0 but not the singularity at  z=1.  Now we choose a contour which
runs around the circle of radius  R>1  except near  z=1  where it angles in
(to avoid 1) and passes through the "saddle points" at  e^{iw}  where
cos(w) = 1- 1/(2n+2).  Near the saddle points it follows the steepest descent
path away from the saddles in both directions.  The contributions from the
sections of the contour away from these saddle points can be bounded away. 
For example on the arc of the circle  |z|=R  the integral is easily bounded
by  C R^{-n}  which is negligible.  The only contributions are from near the
saddle points.  The saddle points where chosen to be the critical points of
the function

	F(z) = z/(z-1) - (n+1) log z

therefore near them our integrand  e^F(z)  can be approximated by a Gaussian
and the contribution of the saddle points follows easily.  (There is some 
hassle to bound away the error terms of course.)  The final result is:

	a_n = sqrt(e/pi) n^{-3/4} cos( 2 sqrt(n) + pi/4) + O(n^{-5/4}).

Note that this agrees very well with the formula derived by Robert Israel and
justifies his assertion that the Taylor series converges at  z=1.

				Richard Stong