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We have to prove the following inequality:

$$\forall n \in \mathbb{N}, n \ge 2, \ \ \Pi_{k=0}^n \binom{n}{k} \le \left(\frac{2^n-2}{n-1}\right)^{n-1}$$

the progress where I am at, is visible in the following picture:

progress

I'm especially confused, since the term (2*((n-1)/n)) has the limit zero when calculated to infinity and the other term diverges to infinity when calculated there. so the first question would be, if these two terms are multiplied, which limit comes out as a result?

the second question would be simply how to continue on from there, or if you know a better way to approach this inequality proof.

I am very thankful for your helpful answers.

Jean Marie
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stanleysearles
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1 Answers1

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By AM-GM inequality $$ \left( {\prod\limits_{k = 0}^n \binom{n}{k} } \right)^{1/(n - 1)} = \left( {\prod\limits_{k = 1}^{n - 1} \binom{n}{k} } \right)^{1/(n - 1)} \le \frac{{\sum\nolimits_{k = 1}^{n - 1} \binom{n}{k} }}{{n - 1}} = \frac{{2^n - 2}}{{n - 1}} $$ provided $n\ge 2$.

Gary
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