# Thread: Why is this like this? Ratio Test on Series

1. ## Why is this like this? Ratio Test on Series When i plug in the (n+1), the bottom isnt 5*8*11...(3(n+1)+2) but it is actually 5*8*11...(3n+2)*(3(n+1)+2)

how come?

2. ## Re: Why is this like this? Ratio Test on Series Originally Posted by lc99  When i plug in the (n+1), the bottom isnt 5*8*11...(3(n+1)+2) but it is actually 5*8*11...(3n+2)*(3(n+1)+2)

how come?
Maybe the (3n+2) is the factor just before (3(n+1)+2)????

In the original expression the next to the last term was 3(n-1) +2. Just because it is not listed doesn't mean it does not exist. Now when YOU replace n with n+1 what do you get??

3. ## Re: Why is this like this? Ratio Test on Series

Hey, when i replaced the n in (3n+2) with n+1, and also the n's in the numerator, i get that the limit is infinite which does not conclude anything about the original series. However, my professor has her answer key and it shows that she added a (3n+2) in the denominator and replaced THAT n with n+1

4. ## Re: Why is this like this? Ratio Test on Series Originally Posted by lc99 However, my professor has her answer key and it shows that she added a (3n+2) in the denominator and replaced THAT n with n+1
Yes! I agree that you should replace ALL the n's with n+1. This includes the last term and the term before the last one.....

5. ## Re: Why is this like this? Ratio Test on Series Originally Posted by lc99  When i plug in the (n+1), the bottom isnt 5*8*11...(3(n+1)+2) but it is actually 5*8*11...(3n+2)*(3(n+1)+2)

how come?
5*8*11...(3(n-1)+2)*(3n+2). Let's just look at the last 2 terms. Now we replace n with n+1 and get (3((n+1)-1)+2)*(3(n+1)+2) = (3(n)+2))*(3(n+1)+2) = (3n+2))*(3(n+1)+2)

6. ## Re: Why is this like this? Ratio Test on Series Originally Posted by lc99  When i plug in the (n+1), the bottom isnt 5*8*11...(3(n+1)+2) but it is actually 5*8*11...(3n+2)*(3(n+1)+2)
how come?
Because this is an alternating series none of the usual test work.
The non alternating part is $a_n=\dfrac{2^n\cdot n!}{5\cdot 8\cdots (3n+2)}$.
If the sequence $a_n$ is decreasing to zero then the series converges.

7. ## Re: Why is this like this? Ratio Test on Series

ahhh. i get it now. thank you! i just never seen the type of series with a sequence like that. and combine thay with convergence tests