Question

Determine whether the sequence converges or diverges. If it converges, find the limit.

(n^2)/√ (n^3+ 9n)

Answer 1

as n gets huge, the limit approaches n^2 / n^(3/2) or n^(1/2)
Looks like it diverges.

Answer 2

Can you show more steps

Answer 3

(n^2)/√ (n^3+ 9n)
= n^2 / √((n^3)(1+9/n^2))
as n gets huge, 1/9^2 goes to zero

Answer 4

let x = (n^2)/√ (n^3+ 9n)
x^2 = n^4/(n^3+ 9n) = n - 9n/(n^3 + 9n)

for n being a whole number , the terms are
1/√10 , 4/√26 , 9/√54, 16/√100, ....
.3162.., .7844..., 1.2247..., 1.6,.....
.3162.. < .7844...< 1.2247...< 1.6 < .....< (if n = 50, 7.058....)
sure looks like the terms are getting bigger, so ....

how about the graph:
www.wolframalpha.com/input/?i=n%5E2%2F%E2%88%9A(n%5E3+%2B+9n)+for+n+%3E+0

Answer 5

I don't know did I follow your steps correctly but I think this is how you did it, right?
(n^2)/√ (n^3+ 9n)
= n^2 / √(((n^3)/(n^3))+(9n/n^3))
=n^2 / √((1+9/n^2))
= n^2 / √(1)
=n^2
=∞ or diverges

Thank you

Answer 6

The graph is help but I need to learn the steps for the test.

Answer 7

when dealing with polynomials and rational functions, as x -> ∞ only the highest power matters. All the lower powers become insignificant in comparison. For example,

(x^3+x-3)/(2x^3-3x^2+5x-9) -> 1/2
because if you divide top and bottom by x^3, you have
(1 + 1/x^2 - 3/x^3) / (2 - 3/x + 5/x^2 - 9/x^3)
All those fractions -> 0 as x->∞ and you are left with just 1/2

That's what I did with your problem. But you made a mistake in your simplification. You cannot just throw away the √n^3 factor
= n^2 / √(((n^3)/(n^3))+(9n/n^3))
= n^2 / √n^3 * √(1+9/n^2)
Now, as x->∞ that becomes
-> n^2/(√n^3*1) = √n