# Prove that a sequence is eventually decreasing.

1. Feb 12, 2010

### The_Iceflash

1. The problem statement, all variables and given/known data

Prove that if a > 0, the sequence $$\frac{a^{n}}{n!}$$ is eventually decreasing for large n.

2. Relevant equations
N/A

3. The attempt at a solution

I know I'm to prove this by finding the index but I'm at a loss on how to find the index on such a sequence as I'm not used to using a variable that's in the sequence.

To warmup I was given a values 110 and 300 to find an index for but with 'a' being in the sequence itself I'm not sure how to do that.

2. Feb 12, 2010

### Gib Z

Try showing that $n! - a^n$ is eventually increasing.

3. Feb 13, 2010

### HallsofIvy

Staff Emeritus
The numerator and denominator are products of n terms but each term in the numerator is a while the terms in the denominator are increasing. The "average" value (geometric average since we are multiplying) in the denominator is $\sqrt[n]{n!}$. Can you show that, for fixed a, there exist an integer, n, such that $\sqrt[n]{n!}> a$?

4. Feb 14, 2010

### The_Iceflash

I don't really follow.

5. Feb 14, 2010

### rsa58

try subtracting the n+1 term of the sequence from the nth term. if this quantitiy is less that zero the the sequence is decreasing. however you can factor out a to the power n. this is greater than zero so it doesn't matter. look at the other factor, if you choose a <N+1 you get the desired result.