Question

can someone help me out with these...stating which test for convergence

Answer 1

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.

Answer 2

So is 5 and 6 Comparison then?

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

haha thanks :)