55 786
Assignments Done
97,2%
Successfully Done
In December 2017
Your physics homework can be a real challenge, and the due date can be really close — feel free to use our assistance and get the desired result.
Be sure that math assignments completed by our experts will be error-free and done according to your instructions specified in the submitted order form.
Our experts will gladly share their knowledge and help you with programming homework. Keep up with the world’s newest programming trends.

Answer on Real Analysis Question 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
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.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question