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
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