# alternate series question

• Apr 17th 2007, 11:02 PM
rcmango
alternate series question
Converges or diverges?

SUM n_infinity (-1)^n(2^n*n!)/(5*8*11*...*(3n +2)

i believe this is an alternating series.

the 5*8*11*.. confuses me.

thankyou for any help.
• Apr 18th 2007, 01:53 AM
Soroban
Hello, rcmango!

Quote:

Converges or diverges?

SUM n_infinity (-1)^n(2^n*n!)/(5*8*11*...*(3n +2)

I would use the Ratio Test

a
n+1 . . . . .2^n+1·(n+1)! . . .5·8·11···(3n+2) . . . .2(n + 1) . . . . 2n + 2
------ . = . ------------------ · ------------------ . = . ---------- . = . ---------
. a
n . . . . .5·8·11···(3n+5) . . . 2^n·n! . . . . . . . . 3n + 5 . . . . . 3n + 5

. . . . . . . . . . . . . . . . . . . . . 2 + 2/n
Divide top and bottom by n: . ---------
. . . . . . . . . . . . . . . . . . . . . 3 + 5/n

Now take the limit as n → ∞

• Apr 18th 2007, 09:37 AM
CaptainBlack
Quote:

Originally Posted by Soroban
Hello, rcmango!

I would use the Ratio Test

an+1 . . . . .2^n+1·(n+1)! . . .5·8·11···(3n+2) . . . .2(n + 1) . . . . 2n + 2
------ . = . ------------------ · ------------------ . = . ---------- . = . ---------
. an . . . . .5·8·11···(3n+5) . . . 2^n·n! . . . . . . . . 3n + 5 . . . . . 3n + 5

. . . . . . . . . . . . . . . . . . . . . 2 + 2/n
Divide top and bottom by n: . ---------
. . . . . . . . . . . . . . . . . . . . . 3 + 5/n

Now take the limit as n → ∞

Now this is overkill, we have proven that the series is absolutely convergent.

We could have proven that it converges (without proving that it is
absolutely convergent) by using the Alternating Series test.

This requires that we show that the absolute value of the terms is
eventually decreasing, and the limit of the terms is 0.

RonL
• Apr 18th 2007, 09:39 AM
CaptainBlack
Quote:

Originally Posted by Soroban
Hello, rcmango!

I would use the Ratio Test

an+1 . . . . .2^n+1·(n+1)! . . .5·8·11···(3n+2) . . . .2(n + 1) . . . . 2n + 2
------ . = . ------------------ · ------------------ . = . ---------- . = . ---------
. an . . . . .5·8·11···(3n+5) . . . 2^n·n! . . . . . . . . 3n + 5 . . . . . 3n + 5

This seems to have lost the alternating signs of the original series, so is
actually

|a_{n+1}/a_n|

and so you have proven absolute convergence

RonL