Calculus II - Infinite Series

1. Sep 18, 2011

GreenPrint

1. The problem statement, all variables and given/known data

Prove that

sigma[n=0,inf] ((-1)^n n)/(n+1)
diverges

2. Relevant equations

3. The attempt at a solution

I'm unsure how to do this

I tried applying the alternating sereis test but when I did so I got
(n/(n+1))' = 1/(n+1)^2
so I can't say that the terms are non increasing and so the conditions are not meant for the alternating series test

I tried applying the root test and got an infinite loop of l hospital's rule (or however it's spelled) because of the square root so I couldn't come up with a conclusion based on that

I tried applying the ratio test and got r = 1 so the test was inconclusive

I don't know how to integrate something like ((-1)^n n)/(n+1)
so I concluded that the integral test was not going to work

Can I just simply say by the divergence test
lim n->inf ((-1)^n n)/(n+1) =/= 0
I'm having a hard time justifying that this is a true statement though...

I'm not experienced with applying comparison tests to alternating series...
thanks for any help

2. Sep 18, 2011

vela

Staff Emeritus
Yes, that's the simplest argument for why the series doesn't converge. Why are you having a hard time justifying the statement?

3. Sep 18, 2011

GreenPrint

Well I'm just having a hard time stating why it's true, I thought maybe because the term (-1)^n is undefined so the limit does not equal zero, but it just seems very vague so I'm having a hard time just like proving it.

4. Sep 18, 2011

vela

Staff Emeritus
That term just causes the sign of the terms to alternate. It's not the reason why the series won't converge. For example, the alternating harmonic series $$\sum_{n=0}^\infty \frac{(-1)^n}{n}$$ converges even though it contains that factor.

What's happening to the values of the sequence$$a_n = (-1)^n \frac{n}{n+1}$$ as n gets large? How does it behave differently than the terms in the alternating harmonic series?

5. Sep 18, 2011

GreenPrint

the lim n->inf n/(n+1) = 1
while lim n-> inf 1/n approaches zero...

so I can say that the series diverges because (-1)^n n/(n+1) near n = infinity alternates between two values 1 and -1 (two different values) so the limit does not equal zero

and lim 1/n approaches zero from both sides as n goes to infinity (the same value) so it converges

is that the idea behind it?

6. Sep 18, 2011

vela

Staff Emeritus
Yes. The limit doesn't exist for the first sequence, so it obviously can't equal 0. Therefore, the series won't converge.

For the second series, the limit of the sequence does exist and equals 0, so the series may converge. (You'd still have to prove it converges by another method if you were trying to answer that question.)

7. Sep 18, 2011

GreenPrint

Hm interesting thanks