Question

Series help.

Answer 1

\[\sum_{n=0}^{\infty}~~\frac{\cos(n\pi)}{2n+7}\]

Answer 2

as n approaches infinity cos will just fluctuate so you want to look at 1/2n+7 which will converse as n approaches infinity

Answer 3

lim n -> inf

1/ 2n+7 = 0

Answer 4

Oh, so you took the limit of the whole thing and then you analyzed it when n approaches infinity?

Answer 5

i think theres a test where you can say the original converges if the 1/2n+7 part converges

Answer 6

cause the original is smaller? I dont remember its been a while

Answer 7

The series, to me, looked like it can be tested by the integral test, but i had a difficult time knowing whether it would be decreasing or not. I don't know for sure if I considered the appropriate test for this series.
Also, when i first did the divergence test (nth term test), like i usually start off with most questions, i didn't get the appropriate result, not a number nor a zero.

Answer 8

https://math.dartmouth.edu/archive/m8w10/public_html/m8l09.pdf

While we have stated the test with 1 n 1, it of course applies if the terms involve
1 n instead (or cos nπ, since this is just a convoluted way to write 1 n). Also, notice
that the Alternating Series Test can not be used to show that a series diverges (see Example
2).

Answer 9

2nd paragraph

Answer 10

cos npi returns 1 or -1 so you need to figure out what test lets you ignore that term

Answer 11

Soon as I see cos(n pi) I usually think of alternating series test ;).

Answer 12

\[\Large \cos(n \pi) = (-1)^n \] ....??!?!

Answer 13

It is an alternating series test. I figured it our after i expanded it xD
\[\sum_{n=0}^{\infty}\frac{\cos(n \pi)}{2n+7}=\frac{1}{7}-\frac{1}{9}+\frac{1}{11}-\frac{1}{13}+...\]

Answer 14

I should start consider expanding.