Answer to Question #3008 in Real Analysis for Sue

Question

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

Answer 1

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
II.
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.