# Thread: help with sum for absolute convergence

1. ## help with sum for absolute convergence

SUM n_infinity (-1)^n+1 (3n/n!)

I believe this is solved using the absolute ratio test.

I see where this becomes lim n -> infinity (3n+1)/(n+1)! / 3n/n!

=

lim n->infinity 3/(n +1) = 0

Is it common for this series to always divide by the original equation? what happened to the (-1)^n+1

i need some help understanding whats going on.

2. Originally Posted by rcmango
SUM n_infinity (-1)^n+1 (3n/n!)

I believe this is solved using the absolute ratio test.

I see where this becomes lim n -> infinity (3n+1)/(n+1)! / 3n/n!

=

lim n->infinity 3/(n +1) = 0

Is it common for this series to always divide by the original equation? what happened to the (-1)^n+1

i need some help understanding whats going on.
you are incorrect with the first part.

it's (3(n+1))/(n+1)! / 3n/n! not (3n+1)/(n+1)! / 3n/n! you forgot the brackets around n+1, that makes the answer totally different. in fact, it makes the solution unknown by the ratio test, since using the ratio test will give a limit of 1.

the (-1)^n+1 doesn't matter since you took absolute values when doing the test, so you assume it is always positive.

use the alternating series test here

3. okay, your saying to use the alternating series to prove its convergent with this part: (-1)^n+1

thus being always positive?

4. Originally Posted by rcmango
okay, your saying to use the alternating series to prove its convergent with this part: (-1)^n+1

thus being always positive?
hmm, i don't really get your question. do you know what the alternating series test is? you pretty much forget about the (-1)^(n+1).

once the terms are strictly decreasing (that is if we treat them as all positive) and the limit of the expression goes to zero as n goes to infinity, then it is absolutely convergent by the alternating series test. and if a series is absolutely convergent, then it's convergent

5. ya, i see now. I was confused by the part were supposed to ignore.

thanks for the help.