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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?

user13
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2 Answers2

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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.

Peter Foreman
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    Yes, thank you. However, I would like to see where is the flow in my intuition with the contradiction method? – user13 Jun 15 '19 at 14:09
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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.

Minus One-Twelfth
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  • Yes. My logic was can’t we take as k the numerator of the highest term on the sequence dependent on the n like n to the power of n or n+1 to the n+1 etc.? – user13 Jun 15 '19 at 14:20
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    $k$ cannot depend on $n$ at all. In general, the definition of a sequence $(a_n)$ being bounded is that there exists some $K > 0$ such that for all $n\in \Bbb{N}$, we have $|a_n| \le K$. Note the order of the "there exists" and the "for all" statements: the "there exists" came before the "for all" part. This means that the same $K$ must hold for all $n$ simultaneously. This is something you will learn if you learn some basics of logic. Perhaps read some answers here: https://math.stackexchange.com/questions/304217/is-forall-x-exists-y-qx-y-the-same-as-exists-y-forall-x-qx-y. – Minus One-Twelfth Jun 15 '19 at 14:23
  • Note: a different statement would be obtained by switching the "for all" and "there exists" orders above (which would give the statement "for all $n\in\Bbb{N}$, there exists some $K>0$ such that $|a_n|\le K$"). In this statement, because the "for all" part comes first, the $K$ would be allowed to depend on $n$. So for this statement, $K$ would be allowed to be $n^n$ and maybe this is the statement you were having in mind. However, changing the order of the "for all" and "there exists" parts of a statement gives us a completely different statement, and this is not the one to use here. – Minus One-Twelfth Jun 15 '19 at 14:28