Prove that a sequence is eventually decreasing.

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Homework Help Overview

The discussion revolves around proving that the sequence \(\frac{a^{n}}{n!}\) is eventually decreasing for large \(n\), given that \(a > 0\). Participants are exploring the behavior of this sequence in the context of factorial growth versus exponential growth.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Some participants attempt to find an index for the sequence, expressing uncertainty about handling the variable \(a\) within the sequence. Others suggest showing that \(n! - a^n\) is eventually increasing as a potential approach. There are discussions about the average value of terms in the numerator and denominator, specifically questioning the existence of an integer \(n\) such that \(\sqrt[n]{n!} > a\). Additionally, one participant proposes examining the difference between consecutive terms of the sequence to establish decreasing behavior.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants are questioning assumptions and definitions related to the sequence, while others are offering potential methods to analyze its behavior. There is no explicit consensus yet, but several lines of reasoning are being actively considered.

Contextual Notes

Participants are grappling with the implications of having a variable \(a\) in the sequence and how it affects their ability to find an appropriate index. There is also mention of specific values (110 and 300) used as examples for finding an index, indicating a potential constraint in the problem setup.

The_Iceflash
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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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Try showing that [itex]n! - a^n[/itex] is eventually increasing.
 
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]?
 
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.
 
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.
 

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