# Sequence problem.

• Jul 12th 2009, 12:28 PM
Hikari
Sequence problem.
For the following sequence, I'm supposed to plot it then guess what value it converges to (if it does). Then, I have to prove that my guess was right.

#1: An = n^3/n!

So, the first few terms of the seq are: 1,4,4.5,2.7,1.04,.3,etc....
When graphed, it looks as if it might converge to zero.

Now, I'm having a bit of trouble understanding the proof.

The solution manual rearranges An to look like this:

(n/n) * (n/(n-1)) * (n/(n-2)) *(n/(n-3))*....*(1/3)*(1/2)*(1)

So far, this makes sense. But then they choose to pull out

n^2/[(n-1)(n-2)(n-3)] for n>=4 and call it Cn, so that 0<An<=Cn. Then Squeeze Theorem can be used.

So why did they choose n>=4? Why not n>=3 or n>=5? Is there another way to do this?
• Jul 12th 2009, 12:33 PM
Jhevon
Quote:

Originally Posted by Hikari
For the following sequence, I'm supposed to plot it then guess what value it converges to (if it does). Then, I have to prove that my guess was right.

#1: An = n^3/n!

So, the first few terms of the seq are: 1,4,4.5,2.7,1.04,.3,etc....
When graphed, it looks as if it might converge to zero.

Now, I'm having a bit of trouble understanding the proof.

The solution manual rearranges An to look like this:

(n/n) * (n/(n-1)) * (n/(n-2)) *(n/(n-3))*....*(1/3)*(1/2)*(1)

you have a typo here. the red n should be a 1 i think

Quote:

So far, this makes sense. But then they choose to pull out

n^2/[(n-1)(n-2)(n-3)] for n>=4 and call it Cn, so that 0<An<=Cn. Then Squeeze Theorem can be used.

So why did they choose n>=4? Why not n>=3 or n>=5? Is there another way to do this?
lets say they said n>=3, what happens if n = 3? division by zero :eek:

they chose the range of n so as to not cause problems with Cn, n <= 3, we have division by zero due to one of the factors (n - 1), (n - 2), or (n - 3)
• Jul 12th 2009, 02:50 PM
Hikari
Oh, you're right about the typo. Sorry. ^^

And thanks! I understand now.