Question

Help determine whether this series converges or diverges

Answer 1

\[\sum_{n=1}^{n=\infty} \frac{ 4 }{ 5\sqrt{n} +4\sqrt[3]{n} }\]

Answer 2

as n->infinity, the expression is equal to C*n^(-1/3), where C is some constant since n^(-1/3) >> n^(-1/2)
now take a lower bounding integral:
integrate(n^(-1/3),{n,1,infinity})
the integral is infinite, therefore the series also approaches infinity and does not converge at a finite value.

Answer 3

is there a way you could explain by using the comparison test?

Answer 4

4/9 * n^(-1/3) < 4/(5sqrt(n)+4cuberoot(n)) for all n
since n^(-1/3) > n^(-1/2)

Answer 5

then just show 4/9 * n^(-1/3) does not converge, therefore 4/(5sqrt(n)+4cuberoot(n)) cannot converge

Answer 6

why did you copare to n^(-1/3) and not n^(-1/2)

Answer 7

actually you're right. it should be n^(-1/2), to minimize the value of the approximation

Answer 8

4/9 * n^(-1/2) < 4/(5sqrt(n)+4cuberoot(n)) for all n

Answer 9

the lower bounding integral will still equal infinity