Question

Alternating Series.
Determine if Sigma (n=1 to infinity) of (-1)^n (2n/15n+1) converges. I know that we take the limit of (2n/15n+1) as n approaches infinity. But; I'm not sure what to do since we get infinity/infinity when we take the limit...

Answer 1

convergence looks at the ration of next/now to see how fast it goes to zero

Answer 2

but it helps if a read it right

Answer 3

we can determine this from the horizontal asymptote i believe

Answer 4

2n/(15n+1) , for larger and larger value of n looks more like:

2n/15n would you agree?

Answer 5

Absolutely! !!

Answer 6

So now we have an (An) and (bn) to try the limit comparison test?

Answer 7

then for large value of n. this limits out to 2/15

Answer 8

we dont even need to do a limit comparison test since this thing doesnt even go to zero

Answer 9

if the limit of the function doesnt go to zero, it diverges

Answer 10

Oooo OK!

Answer 11

Could we also use L'Hopital's rule since we get an indeterminate form of infinity/infinity?
Which would then give us that 2/15?

Answer 12

no need, but yes, you could if you got bored enough to carry it on :)

Answer 13

YES!

Answer 14

I think it might be a little different since it is an alternating sequence. Consider this.\[\sum_{n=1}^{\infty}(-1)^n (2n/(15n+1))=\sum_{k=1}^{\infty}(-\frac{4k-2}{30k-14}+\frac{4k}{30k+1})\]

Answer 15

The first term is the 2k-1 term, corresponding with the odd values of n (which are all negative), while the second term is the 2k term that corresponds to the even values of n.

Answer 16

Continuing....\[=\sum_{k=1}^{\infty})\frac{(2-4k)(30k+1)+4k(30k-14)}{(30k-14)(30k+1)}\]

Answer 17

\[=\sum_{k=1}^{\infty}\frac{-120k^2+56k+2+120k-56k}{900k^2-390k-14}=\sum_{k=1}^{\infty}\frac{2}{900k^2-390k-14}\]

Answer 18

Pretty sure that converges.

Answer 19

Certainly, if the series is not alternating, it diverges, but since it alternates it will converge.

Answer 20

@squirefreak Take a look at my work on your problem. I came to a different conclusion.