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

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

Mathematics
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\[\sum_{n=2}^{\infty} n / \sqrt{n ^{3-1}}\]
Is that \(n^{3-1}\) inside the radical?
yes ; those are all in the denominator. and the n is in the numerator

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So basically your term is \(n/n=1\)?. Then the series diverges.
oh,, no........ (n^3) -1
any idea as to what I should compare it to?
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 \(\lim_{n\to\infty} a_n/b_n\).
how do I compute?
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.
dont I change the n's to x's?
Yes, that would be better :). Since we need continuity of the quare root function here.
would it just be infinity over infinity? .. I am missing something
you can use L'hospital rule, or divide the expression inside the radical by x^3 (top and bottom).
\[x ^{1/2} + 1/2x ^{-1/2} / 1/2(x ^{3} -1)^{-1/2}\]
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 \(x\to\infty\)

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