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Prove the sequence is $a_{n} = \frac{n^{n}}{n!}$ not bounded above.

This is exercise 1.4.2 in Introduction to Analysis by Arthur P. Mattuck.

The hint given in the book is "Show that $a_{n} > n$".

I attempted to show this using induction.

Basic Step

$a_{1} = \frac{1^{1}}{1} = 1 \not > 1$

$a_{2} = \frac{2^{2}}{2} = 2 \not > 2$

$a_{3} = \frac{3^{3}}{3} = \frac{9}{2} > 3$

Inductive Step

Show $\frac{n^{n}}{n!} > n \implies \frac{(n+1)^{(n+1)}}{(n+1)!} > n+1$

This is where I get stuck.

I reduced the sequence so $\frac{(n+1)^{(n+1)}}{(n+1)!} = \frac{(n+1)^{n}}{n!}$.

I figure the next step is to show $\frac{(n+1)^{n}}{n!} > \frac{n^{n}}{n!} + 1$, but I am not sure how to get there.

I also looked at this question since it was working with a similar equation, but I wasn't able to derive anything out of the discussion that could lead to an answer.


Side Question

Does this need to be solved using induction, or are there other methods of directly proving this result? Is induction a clumsy or unwieldy tool for this problem?

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Note that $a_3=\tfrac{3^3}{3!}=\tfrac{9}{2}$. – Servaes Jun 24 '14 at 6:32
You could use the binomial expansion of $(n+1)^n$ to find $$(n+1)^n=n^n+n\cdot n^{n-1}+\ldots+1,$$ from which the desired inequality follows. – Servaes Jun 24 '14 at 6:35
up vote 12 down vote accepted

$$a_n=\frac{n\cdot n\cdots n}{1\cdot 2\cdots n}=\frac{n}{1}\frac{n}{2}\cdots\frac{n}{n}>\frac{n}{1}=n$$

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