# Summation of Divergent Series

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In physics one often needs to sum divergent series. Here we give a simple example of how one can estimate the limit of x to infinity of a function f(x) given by its Taylor expansion around x = 0 when its radius of convergence is only 1.

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Over 1300 words of explained calculations, with a reference, to show how to estimate the limit of a function given by its Taylor expansion with a convergence radius of 1.

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An unknown function for which we only know the first few terms of a series expansion can often be computed quite accurately for values where the expansion diverges. The approximation can be improved if we have some information about the asymptotic behavior of the function. We'll illustrate this using the following example. A function f(x) has a series expansion around zero given by:

f(x) = x - x^3/3 + x^5/5 - x^7/7 + ... (1)

Of course, given this form we can guess that the nth term is (-1)^(n+1)x^(2n-1)/(2n-1) and that f(x) = arctan(x). In what follows we'll pretend that we don't know anything about f(x) beyond the first 4 terms of the series expansion given in (1), and that the function tends algebraically to a limit at infinity. We want to estimate this limit. This seems an impossible task as the radius of convergence of the series is only 1, for even small x the partial sums start to oscillate wildly. E.g. for x = 3 the partial sums are 3, -6, 42.6, -269.83,..., so it seems doubtful that one could say something sensible about the limit of x to infinity. However, it is possible to improve the convergence using conformal mappings. A practical method to estimate the behavior at infinity works as follows. We apply the conformal transform:

x = p z/(1-z^2) (2)

Where p is a parameter that has to be chosen in some optimal way. The functional form is chosen such that z=0 corresponds to x = 0, allowing us to expand the function f in powers of z. Then z = 1 corresponds to the limit of x to infinity, if this lies inside the radius of convergence, then we can get an accurate estimate. This need not be the case, it is enough if the terms become smaller, the radius of convergence can e.g. be zero for the resummed series to be useful.

Often one takes the denominator to be 1-z, but here we choose this to be 1-z^2 as that is more appropriate for an odd function. It turns out that the optimal way to choose the parameter p ...

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