Proof of factorial inequality concerning fractions

Question

I'm having trouble with a proof, with the case $n>2$.

THEOREM: For every natural number $n∈N$ where $n≠2$, $∑_{i=1}^ni≤n!$

Let us simplify the statement.

$$\begin{alignat*}{2}\frac{n(n+1)}{2}&≤n!\\ \frac{ \frac{n(n+1)}{2} }{n}&≤\frac{n!}{n}\\ \frac{n+1}{2}&≤(n-1)!\\ \frac{ \frac{n+1}{2} }{n+1}&≤\frac{(n-1)!}{n+1}\\ \frac{1}{2}&≤\frac{(n-1)!}{n+1}\end{alignat*}$$

Case $n=1$:$\ldots$

Case $n=2$:$\ldots$

Case $n>2$:

To represent a natural number n>2, one can take another natural number $k$ and add two to it. Then: $$\frac{1}{2}≤\frac{(k+1)!}{(k+3)}$$

$\ldots?$

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Now, I'm stuck for the last case. I have no idea where to go from there. Any hints?

Answer 1

The case $n=1$ is just $1\le 1$. The case $n=2$ is excluded from the theorem, for $n=3$ we get $6\le 6$.

Now proceed by induction: Assuming $n>2$: $$\sum_{i=1}^{n+1} i = \sum_{i=1}^n i + n + 1 \le n! + n + 1 \stackrel{(\ast)}\le n! (n+1) = (n+1)!$$ In $(\ast)$ we used that $n>2$ so $n+1\le n\cdot n!$ and thus $n! + n+1 \le (n+1)!$

Answer 2

Here is an alternative proof (by induction).

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First, show that this is true for $n=3$:

$\sum\limits_{i=1}^{3}i\leq{3!}$

Second, assume that this is true for $n$:

$\sum\limits_{i=1}^{n}i\leq{n!}$

Third, prove that this is true for $n+1$:

$$\sum\limits_{i=1}^{n+1}i=\left(\color{red}{\sum\limits_{i=1}^{n}i}\right)+(n+1)\leq\color{red}{n!}+(n+1)\leq n!\times(n+1)=(n+1)!$$

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Please note that the assumption is used only in the part marked red.