Different proof for a inequality

Question

I have to prove the following

$$\left(\frac{k}{e} \right)^{k-1} \leq (k-1)!$$

without using $(1+\frac{1}{k})^k < e, \forall k \in \mathbb{N}$. I've tried using the following argument:

$$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} , \forall x\in \mathbb{R}$$

therefore $$e^x > \frac{x^n}{n!}, \forall x > 0 \forall n \in \mathbb{N}$$

this also means $$e^{k-1} > \frac{(k-1)^{k-1}}{(k-1)!} $$

transposing the inequality gives us: $$ \left(\frac{(k-1)}{e} \right)^{k-1}< (k-1)!$$

thats almost, what i want to prove.. is there a way to use this trick and finish the prove correctly ?

Answer 1

For $k=1$, we have $$ (k-1)!=1=\left(\frac ke\right)^{k-1}\tag1 $$

For $k\ge2$, $$ \begin{align} e^k &\gt\color{#C00}{\overbrace{\frac{k^{k+1}}{(k+1)!}}^{\frac{k^{k-1}}{(k-1)!}\frac k{k+1}}}+\overbrace{\quad\frac{k^k}{k!}\quad}^\frac{k^{k-1}}{(k-1)!}+\frac{k^{k-1}}{(k-1)!}+\color{#090}{\overbrace{\frac{k^{k-2}}{(k-2)!}}^{\frac{k^{k-1}}{(k-1)!}\frac{k-1}k}}\tag{2a}\\ &=\left(2+\color{#C00}{\frac k{k+1}}+\color{#090}{\frac{k-1}k}\right)\frac{k^{k-1}}{(k-1)!}\tag{2b}\\ &\gt e\frac{k^{k-1}}{(k-1)!}\tag{2c} \end{align} $$ Explanation:
$\text{(2a):}$ use four terms from the Taylor series
$\text{(2b):}$ gather factors
$\text{(2c):}$ the terms in the parentheses are $\ge\frac{19}6\gt e$

Thus, for $k\ge1$, $$ (k-1)!\ge\left(\frac ke\right)^{k-1}\tag3 $$

Answer 2

Sketch $x\geq 3$:

We show :

$$\ln\left(\left(x-1\right)!\right)-\left(x-1\right)\ln\left(\frac{x}{e}\right)\geq 0$$

For that we rewrite the inequality as :

$$\sum_{i=1}^{n-1}\ln\left(n-i\right)-\left(n-1\right)\ln\left(\frac{n}{e}\right)\geq 0$$

Now we add a term as :

$$\left(\sum_{i=1}^{n-1}(k-i)/(k-i)\cdot\ln\left(n-i\right)\right)-\left(n-1\right)\ln\left(\frac{n}{e}\right)\geq 0$$

So the function :

$$f(i)=(k-i)\ln(n-i)$$ is convex for $2n\geq k+i$ so we can apply weighted Jensen's inequality using bound for the harmonic sum .

So we are done .