Math2400
  • Math2400
can someone help me out with these...stating which test for convergence
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Math2400
  • Math2400
anonymous
  • 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.
Math2400
  • Math2400
So is 5 and 6 Comparison then?

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anonymous
  • 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.
anonymous
  • anonymous
So for number 5, for example, this would be a perfect comparison: \[\frac{ \sin^{2}(5n) }{ n^{2} } \le \frac{ 1 }{ n^{2} }\]
Math2400
  • Math2400
hmm for some reason it says one of them is wrong><
anonymous
  • 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.
anonymous
  • anonymous
So yeah, it actually just all reduces to a p-series. My bad for not noticing.
anonymous
  • anonymous
\[\sum_{n=1}^{\infty}\frac{ (-1)^{n}\ln(e^{n} )}{ n^{5}\cos(n\pi) } = \sum_{n=1}^{\infty}\frac{ (-1)^{n}n }{n^{5}(-1)^{n} } = \sum_{n=1}^{\infty}\frac{ 1 }{ n^{4} }\]
Math2400
  • Math2400
haha thanks :)

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