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^{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} =0

II.

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.

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

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