Question

Testing to see if (-1^{n+1} * n^2)/(n+1)^2 is absolutely convergent, conditionally convergent or divergent, I proved it is not absolutely convergent, what can I do to show i is not conditionally convergent?

Answer 1

\[\sum_{n=1}^{\infty} \frac{ (-1)^{n+1} n^2 }{ (n+1)^2 }\]

Answer 2

How about testing the sequence, see if it's convergent to zero at all?

Answer 3

I honestly don't know what to do to test divergence of the series when It is alternating because the alternating series test is only to show convergence

Answer 4

Okay, you've shown it's not absolutely convergent, right?
To show it's not (even) conditionally convergent, you have to prove that the alternating series is divergent....

A simple way to do that is to find this limit...
\[\Large \lim_{n\rightarrow\infty}\frac{(-1)^{n+1}n^2}{(n+1)^2}\]

Answer 5

well that limit would be 1, and negative 0ne for different values of n. so you can use the nth term test for divergence even though it is alternating?

Answer 6

No... this sequence is divergent, isn't it?

Answer 7

No you can't use the nth term test for divergence? is that what you mean? cause my original question is how can you show this is divergent?

Answer 8

Yeah, well, if the sequence which generates a series is divergent, then the series must also be divergent.

Answer 9

it seems as though you are applying the nth term testt which says:\[\lim_{n \rightarrow \infty} A_{n}\neq0\] then the series \[\sum_{n=1}^{\infty}A _{n}\] diverges as well

Answer 10

can you do that with alternating series?

Answer 11

It's kind of fundamental for series.... regardless of what kind of series it is, if the sequence is divergent, then the series is also divergent. That's easily proved...

Answer 12

ok I am at the end of a very long study session so I think it is all blurring thank you for your help

Answer 13

No problem :)