Prove that a sequence is eventually decreasing.

In summary: I don't really follow.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.In summary, the sequence \frac{a^{n}}{n!} is eventually decreasing for large n.
  • #1
The_Iceflash
50
0

Homework Statement



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


Homework Equations


N/A


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.
 
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  • #2
Try showing that [itex]n! - a^n[/itex] is eventually increasing.
 
  • #3
The_Iceflash said:

Homework Statement



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


Homework Equations


N/A


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.

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 [itex]\sqrt[n]{n!}[/itex]. Can you show that, for fixed a, there exist an integer, n, such that [itex]\sqrt[n]{n!}> a[/itex]?
 
  • #4
HallsofIvy said:
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 [itex]\sqrt[n]{n!}[/itex]. Can you show that, for fixed a, there exist an integer, n, such that [itex]\sqrt[n]{n!}> a[/itex]?

I don't really follow.
 
  • #5
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.
 

1. What is a sequence?

A sequence is a list of numbers that follow a specific pattern or rule. It can be finite or infinite in length.

2. How can you prove that a sequence is eventually decreasing?

To prove that a sequence is eventually decreasing, you must show that there exists a point in the sequence where all subsequent terms are smaller than the previous term.

3. What are the different methods for proving that a sequence is eventually decreasing?

There are several methods for proving that a sequence is eventually decreasing, including induction, direct proof, and contradiction.

4. What is the importance of proving that a sequence is eventually decreasing?

Proving that a sequence is eventually decreasing is important because it allows us to make predictions about the behavior of the sequence and determine its long-term behavior.

5. Can a sequence be eventually decreasing without being strictly decreasing?

Yes, a sequence can be eventually decreasing without being strictly decreasing. This means that there may be some terms in the sequence that are equal, but overall, the sequence is still decreasing.

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