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Find if divergent/convergent:
a(sub n)=(n^2/sqr(n^3+4n)), if convergent, find the limit.

User To Kra
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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
User Nicola Mori
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