Question

using the limit comparison test to prove convergence or divergence of the infinite series.....

Answer 1

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

Answer 2

Is that \(n^{3-1}\) inside the radical?

Answer 3

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

Answer 4

So basically your term is \(n/n=1\)?. Then the series diverges.

Answer 5

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

Answer 6

any idea as to what I should compare it to?

Answer 7

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 \(\lim_{n\to\infty} a_n/b_n\).

Answer 8

how do I compute?

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

\[x ^{1/2} + 1/2x ^{-1/2} / 1/2(x ^{3} -1)^{-1/2}\]

Answer 15

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 \(x\to\infty\)