Question

How would I test if this series converges or not?

Answer 1

\[\sum_{1}^{\infty} (-1)^(n-1) (2n/4n^2+1)\]

Answer 2

yeah that doesnt look right..its (-1) to the power n-1

Answer 3

\[\sum_{n=1}^{\infty} (-1)^{n-1} \frac{2n}{4n^2+1}\]

Answer 4

Yes. How did you get that to come out right

Answer 5

\sum_{n=1}^{\infty} (-1)^{n-1} \frac{2n}{4n^2+1}

Answer 6

I havent been able to figure most of it out yet somehow

Answer 7

use the alternating series test

Answer 8

How do I change it though?does it have to be n+1?

Answer 9

no...it doesn't

Answer 10

not sure how this would work then..I have not really done one of these yet just know about it

Answer 11

you need to show that the sequence alternates, is decreasing and that its limit is zero

Answer 12

Alright. i see it alternates obviously.. and decreasing becauset he 4n^2 so I just find the limit of that?which is 0? so I know it converges?

Answer 13

You need to find the limit of \[2n \over 4n^2+1\]as \(n\) goes to infinity.

Answer 14

Okay so because that limit is 0, we know the series will converges. ahh makes sense. so does the (-1)^n-1 even matter? My professor said it had to be n+1. What about n, or n-1? Is that the same?

Answer 15

I'll use x for multiplication, sorry for the confusing notation.

(-1)^(n+1)=(-1)^(2+n-1)=(-1)^2x(-1)^(n-1)=1x(-1)^(n-1)=(-1)^(n-1)

Answer 16

If you used n instead of n-1 every term would change sign. The series would still converge but the limit would be -1 times the other one.

Answer 17

So if you just use n, it doesnt change anything? I have one with just n I was working on

Answer 18

Using n instead of n-1 is the same as multiplying the whole series with -1.

So the limit is multiplied by -1. But whether the series converges or diverges doesn't change.

Answer 19

Alright I think that helps, thanks