## cherrilyn 5 years ago using the limit comparison test to prove convergence or divergence of the infinite series.....

1. cherrilyn

$\sum_{n=2}^{\infty} n / \sqrt{n ^{3-1}}$

2. watchmath

Is that $$n^{3-1}$$ inside the radical?

3. cherrilyn

yes ; those are all in the denominator. and the n is in the numerator

4. watchmath

So basically your term is $$n/n=1$$?. Then the series diverges.

5. cherrilyn

oh,, no........ (n^3) -1

6. cherrilyn

any idea as to what I should compare it to?

7. watchmath

Ok. The idea is just first to ignore all the constant. If we do that then the term is somehow looks like $$n/\sqrt{n^3}=1/\sqrt{n}$$. So we use $$b_n=1/\sqrt{n}$$. Now the series with $$b_n$$ as it terms is divergent since it is a p-series with p <1. Now you just need to compute $$\lim_{n\to\infty} a_n/b_n$$.

8. cherrilyn

how do I compute?

9. watchmath

So you want to compute $\lim_{n\to\infty}\frac{n}{\sqrt{n^3-1}}\cdot \frac{\sqrt{n}}{1}=\lim_{n\to\infty}\sqrt{\frac{n^3}{n^3-1}}.$You should be able to continue from there.

10. cherrilyn

dont I change the n's to x's?

11. watchmath

Yes, that would be better :). Since we need continuity of the quare root function here.

12. cherrilyn

would it just be infinity over infinity? .. I am missing something

13. watchmath

you can use L'hospital rule, or divide the expression inside the radical by x^3 (top and bottom).

14. cherrilyn

$x ^{1/2} + 1/2x ^{-1/2} / 1/2(x ^{3} -1)^{-1/2}$

15. watchmath

hmm I meant like this $\sqrt{\frac{x^3/x^3}{(x^3-1)/x^3}}=\sqrt{\frac{1}{1-\frac{1}{x^3}}}\to \sqrt{1}$as $$x\to\infty$$