Find if divergent/convergent:
a(sub n)=(n^2/sqr(n^3+ if convergent, find the limit.

We are given the equation an = (n^2/ sqrt(n^3+4n)) and asked to determine if the function is divergent or convergent. In this case, we find the limit of the function as n approaches infinity. 

an =  (n^2/ sqrt(n^3+4n))
lim (n to infinity ) = infinity / infinty: ;indeterminate

Using L'hopitals rule, we derive
lim (n to infinity )  = 2 n / 0.5* ( n^3+4n) ^-0.5 * (3 n2 +4) : infinity / infinity 

again, we derive

lim (n to infinity )  = 2 (0.25) (( n^3+4n) ^-0.5))*(3 n2 +4) / 0.5* ( 6n + 4) :infinity / infinity 
 
again, 
lim (n to infinity )  = 2 (0.25) (6n + 4) / 0.5* ( 6)* 0.5 (( n^3+4n) ^-0.5)) 

this goes on and the function is divergent