1) Proven that by all n ∈ N^∗ 2!4!..(2n)!≥((n+1)!)^n 2) Proven by recurring that Σ


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

$1) P r o v e n t h a t b y a l l n \in N^{*}$ $2! 4! . . (2 n)! ⩾ {((n + 1)!)}^{n}$ $2) P r o v e n b y r e c u r r i n g t h a t$ $\sum_{p = 1}^{n} p p! = (n + 1)! - 1$
	Answered by puissant last updated on 03/Nov/21
	$2) Σ_(p=1) ^n pp! = Σ_(p=1) ^n {(p+1)−1}p! = (p+1)!−p! = 2!−1!+3!−2!+.....+n!−(n−1)!+(n+1)!−n! = (n+1)!−1..$
	$2)$ $\underset{p = 1}{\sum^{n}} p p! = \underset{p = 1}{\sum^{n}} {(p + 1) - 1} p! = (p + 1)! - p!$ $= 2! - 1! + 3! - 2! + . . . . . + n! - (n - 1)! + (n + 1)! - n!$ $= (n + 1)! - 1 . .$
	Answered by puissant last updated on 03/Nov/21

	$1)$ $★ 2! ⩾ {((1 + 1)!)}^{1} \to 2! ⩾ 2! ✓$ $★ 2! 4! ⩾ {((2 + 1)!)}^{2} \to 48 ⩾ 36 ✓$ $\underset{k = 1}{\prod^{n}} (2 k)! ⩾ {((n + 1)!)}^{n}$ $\Rightarrow \underset{k = 1}{\prod^{n}} (2 k)! (2 n + 2)! ⩾ {((n + 1)!)}^{n} (2 n + 2)!$ $\Rightarrow \underset{k = 1}{\prod^{n + 1}} (2 k)! ⩾ {((n + 1)!)}^{n + 1} \underset{k = n}{\prod^{2 n}} (k + 2) ⩾ {((n + 1)!)}^{n + 1}$ $\Rightarrow \underset{k = 1}{\prod^{n + 1}} (2 k)! ⩾ {((n + 1)!)}^{n + 1} ✓$ $. . . . . . . . . L e p u i s s a n t . . . . . . . . . . . .$
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