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
