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Question Number 158340 by LEKOUMA last updated on 02/Nov/21

1) Proven that by all n ∈ N^∗ 2!4!..(2n)!≥((n+1)!)^n 2) Proven by recurring that Σ_(p=1) ^n pp!=(n+1)!−1

$$\left.\mathrm{1}\right)\:{Proven}\:{that}\:{by}\:{all}\:{n}\:\in\:{N}^{\ast} \\ $$$$\:\mathrm{2}!\mathrm{4}!..\left(\mathrm{2}{n}\right)!\geqslant\left(\left({n}+\mathrm{1}\right)!\right)^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{Proven}\:{by}\:{recurring}\:{that}\: \\ $$$$\sum_{{p}=\mathrm{1}} ^{{n}} {pp}!=\left({n}+\mathrm{1}\right)!−\mathrm{1} \\ $$

Answered by puissant last updated on 03/Nov/21

2) Σ_(p=1) ^n pp! = Σ_(p=1) ^n {(p+1)−1}p! = (p+1)!−p! = 2!−1!+3!−2!+.....+n!−(n−1)!+(n+1)!−n! = (n+1)!−1..

$$\left.\mathrm{2}\right) \\ $$$$\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}{pp}!\:=\:\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\left\{\left({p}+\mathrm{1}\right)−\mathrm{1}\right\}{p}!\:=\:\left({p}+\mathrm{1}\right)!−{p}! \\ $$$$=\:\mathrm{2}!−\mathrm{1}!+\mathrm{3}!−\mathrm{2}!+.....+{n}!−\left({n}−\mathrm{1}\right)!+\left({n}+\mathrm{1}\right)!−{n}! \\ $$$$=\:\left({n}+\mathrm{1}\right)!−\mathrm{1}.. \\ $$

Answered by puissant last updated on 03/Nov/21

1) ★ 2! ≥ ((1+1)!)^1 → 2! ≥ 2! ✓ ★ 2!4! ≥ ((2+1)!)^2 → 48 ≥ 36 ✓ Π_(k=1) ^n (2k)! ≥ ((n+1)!)^n ⇒ Π_(k=1) ^n (2k)!(2n+2)! ≥ ((n+1)!)^n (2n+2)! ⇒ Π_(k=1) ^(n+1) (2k)! ≥ ((n+1)!)^(n+1) Π_(k=n) ^(2n) (k+2)≥((n+1)!)^(n+1) ⇒ Π_(k=1) ^(n+1) (2k)! ≥ ((n+1)!)^(n+1) ✓ .........Le puissant............

$$\left.\mathrm{1}\right) \\ $$$$\bigstar\:\mathrm{2}!\:\geqslant\:\left(\left(\mathrm{1}+\mathrm{1}\right)!\right)^{\mathrm{1}} \:\rightarrow\:\mathrm{2}!\:\geqslant\:\mathrm{2}!\:\checkmark \\ $$$$\bigstar\:\mathrm{2}!\mathrm{4}!\:\geqslant\:\left(\left(\mathrm{2}+\mathrm{1}\right)!\right)^{\mathrm{2}} \:\rightarrow\:\mathrm{48}\:\geqslant\:\mathrm{36}\:\checkmark \\ $$$$\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{2}{k}\right)!\:\geqslant\:\left(\left({n}+\mathrm{1}\right)!\right)^{{n}} \: \\ $$$$\Rightarrow\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{2}{k}\right)!\left(\mathrm{2}{n}+\mathrm{2}\right)!\:\geqslant\:\left(\left({n}+\mathrm{1}\right)!\right)^{{n}} \left(\mathrm{2}{n}+\mathrm{2}\right)! \\ $$$$\Rightarrow\:\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\prod}}\left(\mathrm{2}{k}\right)!\:\geqslant\:\left(\left({n}+\mathrm{1}\right)!\right)^{{n}+\mathrm{1}} \underset{{k}={n}} {\overset{\mathrm{2}{n}} {\prod}}\left({k}+\mathrm{2}\right)\geqslant\left(\left({n}+\mathrm{1}\right)!\right)^{{n}+\mathrm{1}} \\ $$$$\Rightarrow\:\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\prod}}\left(\mathrm{2}{k}\right)!\:\geqslant\:\left(\left({n}+\mathrm{1}\right)!\right)^{{n}+\mathrm{1}} \:\checkmark \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.........\mathscr{L}{e}\:{puissant}............ \\ $$

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