## liliy 3 years ago absolute convergence , converge or diverge: n^2/2^n. sumation going from n=1 until infinity

1. anhkhoavo1210

$\sum_{n=1}^{\infty} {\frac{n^2}{2^n}}$

2. anhkhoavo1210

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

3. liliy

how?

4. anhkhoavo1210

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

5. anhkhoavo1210

with $u_n=\frac{n^2}{2^n}$ here

6. liliy

wats sr?

7. liliy

i dont understand the formula ur using

8. anhkhoavo1210

this is d'Alembert test

9. anhkhoavo1210

if this limit < 1, the series is converge

10. liliy

neverheard of that...aight

11. anhkhoavo1210

what is the test you studied?

12. liliy

we did power, ratio?

13. anhkhoavo1210

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

14. liliy

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

15. anhkhoavo1210

yes it is

16. liliy

so how do u write it out?

17. anhkhoavo1210

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

18. liliy

howo does thatcancel out?

19. anhkhoavo1210

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

20. liliy

how do you get that second part of the limit?

21. anhkhoavo1210

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

22. liliy

|dw:1336454779193:dw|

23. anhkhoavo1210

It means $\frac{(n+1)^2}{2^{n+1}}.\frac{2^n}{n^2}$ That's Ok?

24. liliy

ya...now wat?

25. anhkhoavo1210

you see $\frac{2^n}{2^{n+1}}=\frac{1}{2}$

26. liliy

o.m.g. yes! wow. sorry im so slow