Question

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

Answer 1

\[\sum_{n=1}^{\infty} {\frac{n^2}{2^n}}\]

Answer 2

Check it with the following test :
\[\lim_{n\to \infty} {\frac{u^{n+1}}{u^n}}\]

Answer 3

how?

Answer 4

sr, it is \[\lim_{n\to \infty} {\frac{u_{n+1}}{u_n}}\]

Answer 5

with \[u_n=\frac{n^2}{2^n}\] here

Answer 6

wats sr?

Answer 7

i dont understand the formula ur using

Answer 8

this is d'Alembert test

Answer 9

if this limit < 1, the series is converge

Answer 10

neverheard of that...aight

Answer 11

what is the test you studied?

Answer 12

we did power, ratio?

Answer 13

It seems to be a ratio test :-? may be differ name, display ratio test here

Answer 14

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

Answer 15

yes it is

Answer 16

so how do u write it out?

Answer 17

\[\lim_{n\to \infty} {\frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}}}=\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}<1\]

Answer 18

howo does thatcancel out?

Answer 19

I don't understand what you say, but if this limit <1, you can deduce this series converges. Moreover, it is absolute converge because \[\frac{n^2}{2^n}>0\]

Answer 20

how do you get that second part of the limit?

Answer 21

\[\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}\]???