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EXAMPLE 9 Discuss the convergence of the sequence below, where n! = 1 . 2 . 3 . ... . n. 4n! an SOLUTION Both numerator and denominator approach infinity as n - co, but here we have no corresponding function for use with I'Hospital's rule (x! is not defined when x is not an integer. ) Let's write out a few terms to get a feeling for what happens to a, as n gets large. a1 4(1 . 2) a, = 2 . 2 4(1 . 2 . 3) a3 3 . 3 . 3 4(1 . 2 . 3 . ... . n) an n . n . n . ... . n It appears from these expressions that the terms are decreasing and perhaps approach 0. To confirm this, observe that 4 2 . 3 . ... . n an n n . n . ... . n Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator. So 0