# 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) (A

Fact n! > sqrt(2*Pi*n) * (n/e)

From here we get that

B=lim(n->inf) ( 1/sqrt(2*Pi*n) * (A*e/n)

lim(n->inf) ( 1/sqrt(2*Pi*n) ) * lim(n->inf) ((A*e/n)

II.

The series is said to converge absolutely if sum( 0 : inf; abs(a_n)) < infinity

As we known sum(0:inf; |A

B= lim(n->inf) (a_n)=lim(n->inf) (A

^{n}/ 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* 0^{n}=0II.

The series is said to converge absolutely if sum( 0 : inf; abs(a_n)) < infinity

As we known sum(0:inf; |A

^{n}| / n!)= sum(0:inf; |A|^{n}/ n!)=exp( |A|) < infinity for all fixed A.Need a fast expert's response?

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