# Struggling with Question: (n/3)^n < n < (n/2)^n for n > 5

• rival95
In summary, the conversation discusses a method for solving a question using Stirling's approximation and proving it through induction. The method involves taking the Log of the inequality and using Stirling's approximation for large enough n. The conversation also hints at a possible solution using an inequality for n+1.
rival95
Trying to solve a question, but having a hard time starting it.

Question:
(n/3)^n < n! < (n/2)^n ; all integers n > 5

I proved the base case already and made the assumption of P(k) and am having difficulty knowing where to start on my proof of:

P(k+1): ( (k+1)/3)^k+1 < (k+1)! < ( (k+1)/2)^k+1

any help is greatly appreciated.

Here's a possible method that is not induction:

(n/2)^n > 1 for n > 2.
(n/3)^n > 1 for n > 3.

So for n > 3 we can take the Log of the inequality.
Log((n/3)^n) < Log(n!) < Log((n/2)^n)

For n large enough (to be determined) we can use Stirling's approximation
Log(n!) ~ (-1 + Log(n)) n + O(1)

Then cancel out the common (positive) factor of n in the inequality and remove the Log(n) term from each side to get
-Log(3) < -1 < -Log(2)
which is true.

All that remains to be shown is that Stirling's approximation is good enough for integer n>5.

Last edited:
Actually, a proof by induction is not too hard either.
But as this is your first post to these forums I suspect it's a homework problem.
So I won't give you the proof yet.

I found it by starting with the n+1 inequality:
((n + 1)/3)^(n + 1) < (n + 1)! < ((n + 1)/2)^(n + 1)
and noting that it implies
1/3 ((n + 1)/n)^n (n/3)^n < n! < 1/2 ((n + 1)/n)^n (n/2)^n

Now you just need to play with the ((n + 1)/n)^n factors.

## 1. What does the notation "n > 5" mean in the given question?

The notation "n > 5" means that the value of n must be greater than 5 in order for the inequality to hold true. Any value less than or equal to 5 would result in an invalid solution.

## 2. Is there a specific method to solve this type of inequality?

Yes, there are multiple methods to solve this type of inequality. One approach is to graph both sides of the inequality and find the intersection point. Another approach is to use logarithms to simplify the inequality and find the solution.

## 3. Can this question be solved algebraically?

Yes, this question can be solved algebraically using basic algebraic manipulations such as factoring, expanding, and rearranging terms. However, it may require more advanced techniques such as logarithms or polynomial division.

## 4. Are there any restrictions on the value of n?

Yes, there are restrictions on the value of n in this inequality. As stated in the question, n must be greater than 5 for the inequality to hold true. Additionally, n must be a positive integer, as negative or non-integer values would result in an invalid solution.

## 5. Can this inequality be solved for values of n less than 5?

No, the given inequality is only valid for values of n greater than 5. Attempting to solve for values less than 5 would result in an invalid solution. However, the inequality can be modified to include values less than 5 by changing the restrictions and adjusting the equation accordingly.

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