1. ## 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?

2. 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

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

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)

3. Oh, you're right about the typo. Sorry. ^^

And thanks! I understand now.