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

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anhkhoavo1210Best ResponseYou've already chosen the best response.1
\[\sum_{n=1}^{\infty} {\frac{n^2}{2^n}}\]
 one year ago

anhkhoavo1210Best ResponseYou've already chosen the best response.1
Check it with the following test : \[\lim_{n\to \infty} {\frac{u^{n+1}}{u^n}}\]
 one year ago

anhkhoavo1210Best ResponseYou've already chosen the best response.1
sr, it is \[\lim_{n\to \infty} {\frac{u_{n+1}}{u_n}}\]
 one year ago

anhkhoavo1210Best ResponseYou've already chosen the best response.1
with \[u_n=\frac{n^2}{2^n}\] here
 one year ago

liliyBest ResponseYou've already chosen the best response.0
i dont understand the formula ur using
 one year ago

anhkhoavo1210Best ResponseYou've already chosen the best response.1
this is d'Alembert test
 one year ago

anhkhoavo1210Best ResponseYou've already chosen the best response.1
if this limit < 1, the series is converge
 one year ago

liliyBest ResponseYou've already chosen the best response.0
neverheard of that...aight
 one year ago

anhkhoavo1210Best ResponseYou've already chosen the best response.1
what is the test you studied?
 one year ago

anhkhoavo1210Best ResponseYou've already chosen the best response.1
It seems to be a ratio test :? may be differ name, display ratio test here
 one year ago

liliyBest 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
 one year ago

liliyBest ResponseYou've already chosen the best response.0
so how do u write it out?
 one year ago

anhkhoavo1210Best ResponseYou've already chosen the best response.1
\[\lim_{n\to \infty} {\frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}}}=\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}<1\]
 one year ago

liliyBest ResponseYou've already chosen the best response.0
howo does thatcancel out?
 one year ago

anhkhoavo1210Best 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 \[\frac{n^2}{2^n}>0\]
 one year ago

liliyBest ResponseYou've already chosen the best response.0
how do you get that second part of the limit?
 one year ago

anhkhoavo1210Best ResponseYou've already chosen the best response.1
\[\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}\]???
 one year ago

anhkhoavo1210Best 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?
 one year ago

anhkhoavo1210Best ResponseYou've already chosen the best response.1
you see \[\frac{2^n}{2^{n+1}}=\frac{1}{2}\]
 one year ago

liliyBest ResponseYou've already chosen the best response.0
o.m.g. yes! wow. sorry im so slow
 one year ago
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