anonymous
  • anonymous
absolute convergence , converge or diverge: n^2/2^n. sumation going from n=1 until infinity
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sum_{n=1}^{\infty} {\frac{n^2}{2^n}}\]
anonymous
  • anonymous
Check it with the following test : \[\lim_{n\to \infty} {\frac{u^{n+1}}{u^n}}\]
anonymous
  • anonymous
how?

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

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