Show that the sequence is not bounded above

Question

I must show that the sequence is not bounded above:

$a_n =\frac{n^n}{n!}$,

I tried to use proof by contradiction: suppose there is some $k$ such that $a_n\le k$, then $n^n \le kn!$, $n*n*n...*n \le k n(n-1)(n-2)...1$ for sure I know that $n*n*n...*n \ge n(n-1)(n-2)...1$ for $n \ge 1$. However, what I wonder can't there be such a big k for which the statement $a_n < kn!$ is true? For example, $k = n^n$. So, where is the contradiction?

Answer 1

You say

"However, what I wonder can't there be such a big $k$ for which the statement $a_n < kn!$ is true? For example, $k=n^n$."

There can be if $k$ is allowed to depend on $n$. However, the $k$ in your assumption must be a constant (independent of $n$). This is because if we assume that $a_n$ is bounded, the definition of this is that there exists $k$ such that $a_n \le k$ for all $n$. So things like $k=n^n$ are not allowed.

Answer 2

As you have written $$\frac{n^n}{n!}=\left(\frac{n}{n}\right)\left(\frac{n}{n-1}\right)\dots\left(\frac{n}{2}\right)\left(\frac{n}{1}\right)\ge n$$ as every term is greater than $1$ except the first term. Hence $a_n$ cannot be bounded.