Induction to show n/(n^n) is less than .5^k

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SUMMARY

The discussion focuses on proving the inequality \(\frac{n!}{n^n} \leq \frac{1}{2}^k\) for all \(n \geq 2\), where \(k\) is defined as the greatest integer less than or equal to \(n/2\). The user successfully established the base case and induction hypothesis but seeks guidance on demonstrating that \(\left(\frac{n}{n+1}\right)^n\) is less than \(\frac{1}{2}\). The user notes that \(\left(1 - \frac{1}{n+1}\right)^n\) is a decreasing sequence and is bounded above by \(\frac{1}{2}\), but requires a rigorous proof for this assertion.

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Homework Statement



Show that [tex]\frac{n!}{n^n} \leq \frac{1}{2}^k[/tex] for all [tex]n \geq 2[/tex], where k is the greatest integer less than or equal to n/2.


Homework Equations


Mathematical Induction


The Attempt at a Solution


I've shown the base case, and I've assumed that the hypothesis is true for the induction step.
Now I want to show it is true for

[tex]\frac{(n+1)!}{(n+1)^(n+1)} \leq (\frac{1}{2})^k[/tex] where k is now the greatest integer less than or equal to (n+1)/2

Currently I have the LHS looking like this:

[tex](\frac{n!}{n^n})\cdot (\frac{n}{n+1})^n[/tex]

The first term is the term from the induction hypothesis, which is good. But I'm lost on where to go with the second term. To obtain the RHS, I need only show that this term is less than a half. It is clear to me that

[tex](\frac{n}{n+1})^n[/tex]

is a decreasing sequence bounded above by 1/2. But how can I actually rigorously show this? I tried to show it was strictly decreasing by induction (so that I could show the n=2 case was an upper bound), but the terms were unnecessarily complicated and it didn't pan out.

Any suggestions?
 
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Here is something which may help:
[tex] \left(\frac{n}{n+1}\right)^{n}=\left(1-\frac{1}{n+1}\right)^{n}[/tex]

You also know that:

[tex] \left(1-\frac{1}{n+1}\right)^{n+1}<\frac{1}{e}<\frac{1}{2}[/tex]

Play around and see what you get.
 

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