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

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cherrilyn
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=2}^{\infty} n / \sqrt{n ^{31}}\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Is that \(n^{31}\) inside the radical?

cherrilyn
 5 years ago
Best ResponseYou've already chosen the best response.0yes ; those are all in the denominator. and the n is in the numerator

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1So basically your term is \(n/n=1\)?. Then the series diverges.

cherrilyn
 5 years ago
Best ResponseYou've already chosen the best response.0oh,, no........ (n^3) 1

cherrilyn
 5 years ago
Best ResponseYou've already chosen the best response.0any idea as to what I should compare it to?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Ok. 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 pseries with p <1. Now you just need to compute \(\lim_{n\to\infty} a_n/b_n\).

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1So you want to compute \[\lim_{n\to\infty}\frac{n}{\sqrt{n^31}}\cdot \frac{\sqrt{n}}{1}=\lim_{n\to\infty}\sqrt{\frac{n^3}{n^31}}.\]You should be able to continue from there.

cherrilyn
 5 years ago
Best ResponseYou've already chosen the best response.0dont I change the n's to x's?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Yes, that would be better :). Since we need continuity of the quare root function here.

cherrilyn
 5 years ago
Best ResponseYou've already chosen the best response.0would it just be infinity over infinity? .. I am missing something

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1you can use L'hospital rule, or divide the expression inside the radical by x^3 (top and bottom).

cherrilyn
 5 years ago
Best ResponseYou've already chosen the best response.0\[x ^{1/2} + 1/2x ^{1/2} / 1/2(x ^{3} 1)^{1/2}\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1hmm I meant like this \[\sqrt{\frac{x^3/x^3}{(x^31)/x^3}}=\sqrt{\frac{1}{1\frac{1}{x^3}}}\to \sqrt{1}\]as \(x\to\infty\)
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