Inequality on $e$

$\Large{ e < \left( \dfrac{(n+1)^{2n+1}}{(n!)^2} \right)^\frac{1}{2n} < e^\alpha }$

Let $n$ be a positive integer. If $\alpha = 1 + \dfrac{1}{12(n+1)}$ , prove that the above expression holds.

#Calculus #Inequalities #Logarithms #ExponentialFunction #ExponentialInequalities

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Satyajit Mohanty What method did you use to solve this question? Also, one of your questions has totally stumped me. Does it have a nice closed form, or do we have to evaluate it numerically? Can you post a solution to that question too? Thanks.

Samuel Jones - 5 years, 9 months ago

@Samuel Jones I'll add the solution to the problem: 300 Followers Problem - Polynomial Differential Reciprocal Summations!

Satyajit Mohanty - 5 years, 9 months ago

And can you please also tell what method did you use for this inequality problem or at least give a hint?

Samuel Jones - 5 years, 9 months ago

@Satyajit Mohanty Sorry to disturb you again, but can you please add a solution to your 300 followers problem?

Samuel Jones - 5 years, 9 months ago

@Samuel Jones – @Samuel Jones - I've added the solution to the problem 300 Followers Problem - Polynomial Differential Reciprocal Summations!. Please check it.

Satyajit Mohanty - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Inequality on \(e\)

Comments