Prove that $(x + 1) (2 n + 1) \geq 6 {(n!)}^{\frac{2}{n}}$ $(x+1)(2n+1)>= 6(n!)^(2/n)$ ?

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Prove that $(x + 1) (2 n + 1) \geq 6 {(n!)}^{\frac{2}{n}}$ $(x+1)(2n+1)>= 6(n!)^(2/n)$ ?

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Answer 1 · 2017-06-22T12:32:21

See below.

Explanation:

Using the Stirling's asymptotic approximation formula

$n! \approx \sqrt{2 π n} {(\frac{n}{e})}^{n}$ $n! approx sqrt(2pi n)(n/e)^n$ we have

$6 {(n!)}^{\frac{2}{n}} \approx 6 {(\sqrt{2 π n} {(\frac{n}{e})}^{n})}^{\frac{2}{n}} = 6 {(2 π n)}^{\frac{1}{n}} {(\frac{n}{e})}^{2}$ $6(n!)^(2/n) approx 6(sqrt(2 pi n)(n/e)^n)^(2/n) = 6(2pi n)^(1/n)(n/e)^2$ then

$(x + 1) (2 n + 1) \geq 6 {(2 π n)}^{\frac{1}{n}} {(\frac{n}{e})}^{2}$ $(x+1)(2n+1) ge 6(2pi n)^(1/n)(n/e)^2$ implies that $x \geq n$ $x ge n$

So the affirmation is true only for $x \geq n$ $x ge n$

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Prove that $(x + 1) (2 n + 1) \geq 6 {(n!)}^{\frac{2}{n}}$ $(x+1)(2n+1)>= 6(n!)^(2/n)$ ?

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Prove that $(x + 1) (2 n + 1) \geq 6 {(n!)}^{\frac{2}{n}}$ $(x+1)(2n+1)>= 6(n!)^(2/n)$ ?