Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

\[\sum_{n=1}^{\infty} {\frac{n^2}{2^n}}\]
Check it with the following test : \[\lim_{n\to \infty} {\frac{u^{n+1}}{u^n}}\]
how?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question