Find if divergent/convergent:

a (sub n) = (n^2/sqr (n^3+4n)), 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