ak=(2^(2k)k!)/k^k
In this problem use the Ratio Test to decide whether the series converges. which is not the series is divergent.
Compute L=lim|av(n+1)/(avn)|and find the numerical value of the limit L
n->infinity
"v" means down
just plug in the numbers.
a_(n+1)/a_n =
2^(2(k+1))(k+1)! / (k+1)^(k+1)
-------------------------------------------------
2^(2k)k! / k^k
= 4(k/(k+1))^k -> 4/e
so the series diverges
To apply the Ratio Test, we need to compute the limit as n approaches infinity of the absolute value of the ratio of consecutive terms in the series av(n+1)/(avn).
Given the series ak=(2^(2k)k!)/k^k, we can calculate the ratio of consecutive terms as follows:
av(n+1)/(avn) = [(2^(2(n+1))((n+1)!))/(n+1)^(n+1)] / [(2^(2n)(n!))/n^n]
= (2^(2(n+1))((n+1)!))/(n+1)^(n+1) * (n^n)/(2^(2n)(n!))
= (2^2(n+1) * 2^2n * n^n * (n+1)!) / ((n+1)^(n+1) * 2^(2n) * n!)
= (4(n+1) * 4n * n^n * (n+1)!) / ((n+1)^(n+1) * 4n * n!)
We can simplify the expression by cancelling out terms:
av(n+1)/(avn) = (4(n+1) * 4n * n^n * (n+1)!) / ((n+1)^(n+1) * 4n * n!)
= 4(n+1) * n^n / (n+1)^n
Taking the limit as n approaches infinity:
L = lim |av(n+1)/(avn)| as n->infinity
L = lim [4(n+1) * n^n / (n+1)^n] as n->infinity
L = lim [4 * (n+1)/ (n+1) * n^n / (n+1)^(n-1)] as n->infinity
L = lim [4 * n^n / (n+1)^(n-1)] as n->infinity
To find the numerical value of the limit L, we can simplify the expression further:
L = lim [4 * (n/n+1)^n] as n->infinity
Since n/n+1 approaches 1 as n approaches infinity, we can rewrite the limit as:
L = 4 * 1^n as n->infinity
L = 4
So, the numerical value of the limit L is 4.