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

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anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_{n=1}^{\infty} {\frac{n^2}{2^n}}\]
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
Check it with the following test : \[\lim_{n\to \infty} {\frac{u^{n+1}}{u^n}}\]
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
sr, it is \[\lim_{n\to \infty} {\frac{u_{n+1}}{u_n}}\]
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
with \[u_n=\frac{n^2}{2^n}\] here
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
i dont understand the formula ur using
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
this is d'Alembert test
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
if this limit < 1, the series is converge
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
neverheard of that...aight
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
what is the test you studied?
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
we did power, ratio?
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
It seems to be a ratio test :? may be differ name, display ratio test here
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
given 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
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
yes it is
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
so how do u write it out?
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
\[\lim_{n\to \infty} {\frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}}}=\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}<1\]
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
howo does thatcancel out?
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
I don't understand what you say, but if this limit <1, you can deduce this series converges. Moreover, it is absolute converge because \[\frac{n^2}{2^n}>0\]
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
how do you get that second part of the limit?
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
\[\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}\]???
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
dw:1336454779193:dw
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
It means \[\frac{(n+1)^2}{2^{n+1}}.\frac{2^n}{n^2}\] That's Ok?
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
ya...now wat?
 2 years ago

anhkhoavo1210 Group TitleBest ResponseYou've already chosen the best response.1
you see \[\frac{2^n}{2^{n+1}}=\frac{1}{2}\]
 2 years ago

liliy Group TitleBest ResponseYou've already chosen the best response.0
o.m.g. yes! wow. sorry im so slow
 2 years ago
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