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Answer to Question #3008 in Real Analysis for Sue

Question #3008
Show that for any number A, the series Summation of (A^n)/n! converges absolutely, conclude that lim as n goes to infinity (A^n)/n! =0
Expert's answer
At first, check condition for elements of the series:
B= lim(n->inf) (a_n)=lim(n->inf) (An / n!)=
Fact n! > sqrt(2*Pi*n) * (n/e)n
From here we get that
B=lim(n->inf) ( 1/sqrt(2*Pi*n) * (A*e/n)n )=
lim(n->inf) ( 1/sqrt(2*Pi*n) ) * lim(n->inf) ((A*e/n)n )= 0* 0n =0
The series is said to converge absolutely if sum( 0 : inf; abs(a_n)) < infinity
As we known sum(0:inf; |An| / n!)= sum(0:inf; |A|n / n!)=exp( |A|) < infinity for all fixed A.

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