Question

Determine whether the series is absolutely convergent, conditionally convergent or divergent.
Sum (from n=1 to infinity) of [ 2*4*6....(2n)] / n! ]

Answer 1

finding a representation for [ 2*4*6....(2n)] in terms of n! would be my first move

Answer 2

lim n-> infinity | | [ 2*4*6... (2n) (2(n+1)) ] / (n+1)! * [ (n!) / (2*4*6 ... 2n) ] |
Is this correct?

Answer 3

Im not sure if I'm supposed to but 2*4*6... (2(n+1)) or 2*4*6... (2n) (2(n+1)) for the numerator of the an+1

Answer 4

to put*

Answer 5

2*4*6... (2n) (2(n+1)) is right, but you can't do much with (n+1)! in the denom with the series written this way

Answer 6

the (n+1)! = (n+1)(n)! right?

Answer 7

yep, I guess you can deal with this without rewriting the numerator

Answer 8

and the n! will cancel out?

Answer 9

right

Answer 10

so i'll be left with | 2n+2 / n+1 |

Answer 11

and lim-> infinity of this = 2+0/ 1+0 = 2 > 1 divergent?

Answer 12

right :)

Answer 13

alright thank you but how do i determine the second part to prove that it's divergent or conditionally convergent?

Answer 14

oh I'm rusty on that, but it seems that my earlier idea would show that this series is just divergent
rewrite [2*4*6*...*(2n)]=2^n(n!)

Answer 15

oh okay, thank you for your help!! :)

Answer 16

if a series is divergent by the ratio test, it is simply divergent I think
welcome!

Answer 17

oh just to note, if its divergent, the non-absolute value has to be divergent in order to be fully divergent. If the non-absolute value is convergent then it's conditionally convergent. :)

Answer 18

So I guess the non-absolute value of this question is the same.. so it'll be divergent overall. :o hope this is right

Answer 19

like without the ratio test and maybe manually. Not too sure

Answer 20

I think you are right since there is not even any sign to consider here
but as I said earlier, if you rewrite the numerator as 2^n(n!) then you realize this is just\[\sum_n^{\infty}2^n\] which is clearly divergent