Series Converg. Hmwk: Determine Convergence of \sum(-1)^n n/(n^p + (-1)^n)

[STEMucator]

Homework Statement

Really tough series to work with.

Determine the convergence ( absolute or conditional ) or divergence of :

$\sum_{n=2}^{∞} \frac{(-1)^n n}{n^p + (-1)^n}$

Homework Equations

?? Series tests?

The Attempt at a Solution

This series is really ugly. I'm not sure how to apply any of the tests I've been given to this particular series.

The ratio test gives me nada, as do Raabe's and Gauss's tests. The n'th root test fails as well as the integral test. The Cauchy Condensation test also yields no results. It's not a geometric series nor can I compare it to anything as is.

The only tests I think which I have left which will be useful is the Alternating series test and the limit form of the comparison test.

I also thought of something which doesn't really involve too much testing. I was thinking I could perhaps separate the positive and negative terms of the series into two other series.

Let :

$\sum u_n = \sum_{n=1}^{∞} \frac{2n}{2^p n^p + 1}$

and

$\sum v_n = \sum_{n=1}^{∞} \frac{2n+1}{(2n+1)^p - 1}$

Then if I show that u_n and v_n both diverge, it would mean that $\sum |a_n|$ diverged and thus $\sum a_n$ would be conditionally convergent.

I'm not sure if I'm approaching this correctly, or if there's something I may not be seeing, but if anyone knows what direction to take it would be appreciated if you could share.

 

[haruspex]

I would break it into cases according to the range of p. E.g. if p > 0 then for n > N, (n^p+(-1)ⁿ)^-1 can be bounded to be within some range of n^-p.

 

[STEMucator]

I would break it into cases according to the range of p. E.g. if p > 0 then for n > N, (n^p+(-1)ⁿ)^-1 can be bounded to be within some range of n^-p.

Hey harup, thanks for the input. Just a note that p is never zero.

I was thinking of something and I would like to see if maybe this will help in solving this.

So we have : $\sum_{n=2}^{∞} \frac{(-1)^n n}{n^p + (-1)^n} = \sum (-1)^na_n$

Now : $\sum (-1)^na_n ≤ \sum_{n=2}^{∞} \frac{(-1)^n n}{n^p - 1} = \sum (-1)^n b_n$

I'm not sure this will help though.

I don't quite understand how to break it into cases like you're suggesting ( For p > 0 and p < 0 ).

 

[haruspex]

Something like this...
(n^p+(-1)ⁿ)^-1 = n^-p(1+(-1)ⁿn^-p)^-1
|(n^p+(-1)ⁿ)^-1 - n^-p| < n^-2p+n^-3p+... < n^-2pα for n > (1-1/α)^-1/p