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#409278 ·published 2007-03-25 15:44 UTC
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#!/usr/bin/python

def f(n):
	if n < 2:
		return 1
	else:
		return n * f(n-1)

x = 1.0		

e = 1.0

for i in range(1,100):
	d = x**i
	n = f(i)
	e += float(d)/float(n)
	
print e