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I was wondering how the remainder of the stirling formular behaves for $ n \rightarrow \infty $, while $n$ is the index of the bernoulli polynomial.

I got a problem either way: if it diverges then how can wolfram alpha print something like:

http://mathworld.wolfram.com/StirlingsSeries.html at (7)

it seems as they wrote for $ n \rightarrow \infty $ but then where is the remainder term? if it doesnt diverge the sterling series in (7) would have to be convergent which isnt true. i was wondering in general how one gets from the euler mclaurin formula to the stirling formula as in (7) the only way i see is when the remainder becomes zero for $ n \rightarrow \infty $.

can you tell me how the remainder term actually behaves for $ n \rightarrow \infty $?

3 Answers 3

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All the series at that wolfram page derive from the first one there, which is identified as an asymptotic series for the Gamma function. Do you know what an asymptotic series is? If not, there's a link on that page that will help you. Basically, asymptotic series don't converge, but the error is less than the first omitted term.

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See this paper. It contains explicit formulas and error bounds for the remainder term.

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As indicated by Gerry the Stirling is only asymptotic but observe that convergent versions of Stirling formula exist as proposed by Wikipedia.