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anonymous
 4 years ago
absolute convergence , converge or diverge:
n^2/2^n. sumation going from n=1 until infinity
anonymous
 4 years ago
absolute convergence , converge or diverge: n^2/2^n. sumation going from n=1 until infinity

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty} {\frac{n^2}{2^n}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Check it with the following test : \[\lim_{n\to \infty} {\frac{u^{n+1}}{u^n}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sr, it is \[\lim_{n\to \infty} {\frac{u_{n+1}}{u_n}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0with \[u_n=\frac{n^2}{2^n}\] here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont understand the formula ur using

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is d'Alembert test

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if this limit < 1, the series is converge

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0neverheard of that...aight

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is the test you studied?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It seems to be a ratio test :? may be differ name, display ratio test here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0given sumation A sub n. let the ratio= lim n goes to infinity of absolute val of A sub n +1 / A sub n then if ratio<1 series Coverges Absloutely ratio>1, diverges ratio=1, try diff test

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so how do u write it out?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n\to \infty} {\frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}}}=\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}<1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0howo does thatcancel out?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't understand what you say, but if this limit <1, you can deduce this series converges. Moreover, it is absolute converge because \[\frac{n^2}{2^n}>0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how do you get that second part of the limit?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}\]???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1336454779193:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It means \[\frac{(n+1)^2}{2^{n+1}}.\frac{2^n}{n^2}\] That's Ok?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you see \[\frac{2^n}{2^{n+1}}=\frac{1}{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0o.m.g. yes! wow. sorry im so slow
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