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

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

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

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

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– Samuel Jones · 4 months, 3 weeks ago

What has this link you posted got to do with the inequality?Log in to reply

@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 · 1 year, 4 months ago

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@Samuel Jones I'll add the solution to the problem: 300 Followers Problem - Polynomial Differential Reciprocal Summations! – Satyajit Mohanty · 1 year, 4 months ago

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– Samuel Jones · 1 year, 4 months ago

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

@Satyajit Mohanty Sorry to disturb you again, but can you please add a solution to your 300 followers problem? – Samuel Jones · 1 year, 4 months ago

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@Samuel Jones - I've added the solution to the problem 300 Followers Problem - Polynomial Differential Reciprocal Summations!. Please check it. – Satyajit Mohanty · 1 year, 4 months ago

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