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

1 answer

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