Proof without induction : $(n+1)! < n(1!+2!+\cdots +n!)$

Question

Prove that :

$(a)\;\; (n+1)! < n\big(1!+2!+\cdots +n!\big)$

$(b)\;\; (n+1)!>(n-1)\Big(n!+(n-1)!+\cdots \cdots +1!\Big)$ $ \forall n \geq 3$

Without using induction method

Attempt: for first part, $(n+1)!=(n+1)\cdot n\cdot (n-1)! = n\Big((n-1)!+n!\Big)$

for second part, $(n+1)!=(n+1)\cdot n\cdot (n-1)(n-2)! = (n-1)\Big((n^2+n)(n-2)!\Big)$

Could someone help me how to go further?

Thanks.

Answer 1

$$n(1!+2!+...+n!)=n(n-1)!\left(\frac{1}{(n-1)!}+\frac{2}{(n-1)!}+...+1+n\right)>$$ $$>n(n-1)!(n+1)=(n+1)!$$ $$(n-1)(1!+2!+...+n!)=(n-1)n!\left(\frac{1}{n!}+...+\frac{1}{n}+1\right)<$$ $$<(n-1)n!(n+1)=(n+1)!$$

Answer 2

Notice that we have $$ 1+\frac1n+\overbrace{\underbrace{\frac1{n(n-1)}}_{\frac{(n-2)!}{n!}}+\underbrace{\frac1{n(n-1)(n-2)}}_{\frac{(n-3)!}{n!}}+\dots+\frac{0!}{n!}}^{\le(n-1)\cdot\frac1{n(n-1)}=\frac1n}\le\frac{n+2}n\tag1 $$ Since the last $n-1$ terms are no greater than $\frac1{n(n-1)}$, their sum is no greater than $\frac1n$. Thus, the sum of all the terms is no greater than $\frac{n+2}n$.

Therefore, $$ \begin{align} \sum_{k=0}^nk! &=n!\left(1+\frac1n+\frac1{n(n-1)}+\dots+\frac{0!}{n!}\right)\tag{2a}\\ &\lt\frac{(n+1)!}{n-1}\tag{2b} \end{align} $$ Explanation:
$\text{(2a)}$: factor $n!$ out front
$\text{(2b)}$: apply $(1)$ and note that $\frac{n+2}n\lt\frac{n+1}{n-1}$ for $n\ge2$

---

$$ \begin{align} \sum_{k=0}^n\frac{k!}{n!} &=1+\frac1n+\overbrace{\color{#C00}{\frac1{n(n-1)}}}^{\frac{(n-2)!}{n!}}+\overbrace{\color{#090}{\frac1{n(n-1)(n-2)}}}^{\frac{(n-3)!}{n!}}+\dots+\color{#00F}{\frac{1!}{n!}+\frac{0!}{n!}}\tag{3a}\\ &\ge\left(1+\frac1n+\color{#C00}{\frac1{n^2}}+\color{#090}{\frac1{n^3}}+\dots+\color{#00F}{\frac1{n^{n-1}}+\frac1{n^n}}\right)+\left(\color{#C00}{\frac1{n(n-1)}-\frac1{n^2}}\right)\tag{3b}\\ &=\frac{n-\frac1{n^n}}{n-1}+\frac{\frac1{n^2}}{n-1}\tag{3c}\\[6pt] &\ge\frac{n}{n-1}\tag{3d} \end{align} $$ Explanation:
$\text{(3a)}$: expand the sum
$\text{(3b)}$: split apart the $k=2$ term (in red)
$\phantom{\text{(3b):}}$ use $\frac{(n-k)!}{n!}\ge\frac1{n^k}$ for the rest
$\text{(3c)}$: use the sum of a geometric series
$\text{(3d)}$: $n\ge2$ $$ \begin{align} \sum_{k=1}^n\frac{k!}{n!} &=1+\frac1n+\color{#C00}{\frac1{n(n-1)}}+\color{#090}{\frac1{n(n-1)(n-2)}}+\dots+\color{#00F}{\frac{1!}{n!}}\tag{4a}\\ &\ge\left(1+\frac1n+\color{#C00}{\frac1{n^2}}+\color{#090}{\frac1{n^3}}+\dots+\color{#00F}{\frac1{n^{n-1}}}\right)+\left(\color{#C00}{\frac1{n(n-1)}-\frac1{n^2}}\right)\tag{4b}\\ &=\frac{n-\frac1{n^{n-1}}}{n-1}+\frac{\frac1{n^2}}{n-1}\tag{4c}\\[6pt] &\ge\frac{n}{n-1}\tag{4d} \end{align} $$ Explanation:
$\text{(4a)}$: expand the sum
$\text{(4b)}$: split apart the $k=2$ term (in red)
$\phantom{\text{(3b):}}$ use $\frac{(n-k)!}{n!}\ge\frac1{n^k}$ for the rest
$\text{(4c)}$: use the sum of a geometric series
$\text{(4d)}$: $n\ge3$

Therefore, for $n\ge2$, $(3)$ says $$ \sum_{k=0}^nk!\ge\frac{n\,n!}{n-1}\tag5 $$ and for $n\ge3$, $(4)$ says $$ \sum_{k=1}^nk!\ge\frac{n\,n!}{n-1}\tag6 $$