## Math2400 one year ago can someone help me out with these...stating which test for convergence

1. Math2400

2. anonymous

This is based on the answer choices you have typed into the boxes: Number 1 is an alternating series. For all integers n, you either have odd multiples of pi, which make the value of cosine -1, or you have even multiples, which make the value of cosine equal to 1. $$\cos(n\pi), n \in \mathbb{N} = (-1)^{n}$$ 5 and 6 aren't p-series. You can compare them to p-series, but they themselves are not p-series. 3 has the (-1)^n term, so it is an alternating series. Unless you found a way to compare it to a geometric or p-series, which seems like it would be hard to do. 2 and 4 seem fine.

3. Math2400

So is 5 and 6 Comparison then?

4. anonymous

Yes, you can compare them to p-series. A couple notes: A p-series will have this form $a \sum_{n=1}^{\infty}\frac{ 1 }{ n^{p} }$ where a is a constant. P-series will not have sin, cos, ln, n!, e^n, nothing like that. This is also an example of something which is NOT a p-series $\sum_{n=1}^{\infty}\frac{ 1 }{ n^{2}+1 }$ Throwing that extra plus 1 in the denominator makes it no longer a p-series. It must only be a constant times a power of n in the denominator. As for dealing with sin and cos, we know they are bounded between -1 and 1. So when you have a sin or a cos, you can always say it is less than or equal to 1. Its a nice comparison to use to help deal with those.

5. anonymous

So for number 5, for example, this would be a perfect comparison: $\frac{ \sin^{2}(5n) }{ n^{2} } \le \frac{ 1 }{ n^{2} }$

6. Math2400

hmm for some reason it says one of them is wrong><

7. anonymous

Ah, I see why, didnt catch it. There is a cos(npi) in the denominator, soit would cancel with the (-1)^n in the numerator.

8. anonymous

So yeah, it actually just all reduces to a p-series. My bad for not noticing.

9. anonymous

$\sum_{n=1}^{\infty}\frac{ (-1)^{n}\ln(e^{n} )}{ n^{5}\cos(n\pi) } = \sum_{n=1}^{\infty}\frac{ (-1)^{n}n }{n^{5}(-1)^{n} } = \sum_{n=1}^{\infty}\frac{ 1 }{ n^{4} }$

10. Math2400

haha thanks :)