Question

determine the series is aabsolutely convergent

Answer 1

A series would be helpful.

Answer 2

conditionally convergent
http://www.wolframalpha.com/input/?i=summation+from+1+to+infinity+%28-1%29%5En+n%5E2%2F3%5En
as well as absolutely convergent too
http://www.wolframalpha.com/input/?i=summation+from+1+to+infinity+n%5E2%2F3%5En

Answer 3

I think it will converge, but I can not calculate the sum.

Answer 4

\[ \sum_{1}^{}(-1)^n n^3/3^n\]

Answer 5

i want that is absolutely convergent

Answer 6

@experimentX It doesn't make sense to say it is conditionally and absolutely convergent since absolute convergence implies conditional convergence.

Answer 7

Somebody do the sum, Please. :D

Answer 8

Yes it does make sense show absolute and conditional convergence separately. That is discussion for a different forum.

Answer 9

You will never have absolute convergence without conditional convergence.

Answer 10

can we solve this series by the ratio test?

Answer 11

of course, you can. but what's the ratio test? hehe

Answer 12

try integral test

lin x-> inf http://www.wolframalpha.com/input/?i=integrate+x%5E2%2F3%5Ex

Answer 13

The ratio test is always easiest for me. An+1 / An

Answer 14

Ohh, that's simple. and easy. maybe quick too.

Answer 15

Just look at the limit as n^2/3^n goes to infinity. Since it's obvious that n^2 goes to infinity much slower than 3^n (check with your calculator if you don't believe me) you'll see the limit goes to 0. That means it converges, hooray.

Answer 16

looks like the ratio test is easier than integral test,

Answer 17

Is Integral test valid because, \[\sum_{n=0}^{k}\frac{n^2}{3^n} < \int_0^k \frac{n^2}{3^n} \]?
That's how I did too @Kainui

Answer 18

http://www.wolframalpha.com/input/?i=lim+n-%3Einf+%28%28n%2B1%29%5E2%2F3%5E%28n%2B1%29%29%2F%28%28n%29%5E2%2F3%5E%28n%29%29

Answer 19

Taking the limit is the test for divergence, and it is not acceptable to use the test for divergence to prove CONVERGENCE.

Limit does not exist, or infinity, or NOT 0 = diverges
limit = 0 Inconclusive via test for divergence

Answer 20

There are only two conditions to satisfy, supplied you know that the alternating series is convergent to begin with. \[0<a _{k+1}\le a_{k}\] and \[\lim_{k \rightarrow \infty} a _{k}=0\]

Answer 21

@Maekin It's acceptable in this case because you already know that the alternating series is convergent to begin with.

Answer 22

@experimentX and 1/3 <1 so the series is absolutely convergent

Answer 23

Exactly. It's the comparison test that makes the limit thingy work and be so awesome.