1-Proven-that-by-all-n-N-2-4-2n-n-1-n-2-Proven-by-recurring-that-p-1-n-pp-n-1-1-

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)

\underset{p = 1}{\sum^{n}} p p! = \underset{p = 1}{\sum^{n}} {(p + 1) - 1} p! = (p + 1)! - p!

= 2! - 1! + 3! - 2! + \dots . . + 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} ✓

\dots \dots \dots L e p u i s s a n t \dots \dots \dots \dots

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