liliy
absolute convergence , converge or diverge:
n^2/2^n. sumation going from n=1 until infinity
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anhkhoavo1210
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\[\sum_{n=1}^{\infty} {\frac{n^2}{2^n}}\]
anhkhoavo1210
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Check it with the following test :
\[\lim_{n\to \infty} {\frac{u^{n+1}}{u^n}}\]
liliy
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how?
anhkhoavo1210
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sr, it is \[\lim_{n\to \infty} {\frac{u_{n+1}}{u_n}}\]
anhkhoavo1210
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with \[u_n=\frac{n^2}{2^n}\] here
liliy
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wats sr?
liliy
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i dont understand the formula ur using
anhkhoavo1210
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this is d'Alembert test
anhkhoavo1210
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if this limit < 1, the series is converge
liliy
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neverheard of that...aight
anhkhoavo1210
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what is the test you studied?
liliy
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we did power, ratio?
anhkhoavo1210
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It seems to be a ratio test :-? may be differ name, display ratio test here
liliy
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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
anhkhoavo1210
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yes it is
liliy
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so how do u write it out?
anhkhoavo1210
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\[\lim_{n\to \infty} {\frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}}}=\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}<1\]
liliy
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howo does thatcancel out?
anhkhoavo1210
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I don't understand what you say, but if this limit <1, you can deduce this series converges. Moreover, it is absolute converge because \[\frac{n^2}{2^n}>0\]
liliy
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how do you get that second part of the limit?
anhkhoavo1210
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\[\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}\]???
liliy
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|dw:1336454779193:dw|
anhkhoavo1210
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It means \[\frac{(n+1)^2}{2^{n+1}}.\frac{2^n}{n^2}\]
That's Ok?
liliy
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ya...now wat?
anhkhoavo1210
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you see \[\frac{2^n}{2^{n+1}}=\frac{1}{2}\]
liliy
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o.m.g. yes! wow. sorry im so slow