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Consider the sequence

$\binom n k /n^k$

where k is a number from $0$ to $n$. I want to prove that this sequence is increasing in $n$. It doesn't have to be strictly increasing since for $k = 0,1$ the sequence is constant. Increasing is enough.

I tried to used Bernoulli inequality among others but I can't prove it.

Thank you for any help.

Increasing with respect to n I.e as n increases or with respect to k I.e as increases.

mathsdiscussion.com
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Luz
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  • I might take a logarithm and apply https://math.stackexchange.com/questions/2674183/derivative-of-binomial-coefficients . – Eric Towers Aug 10 '19 at 00:58

2 Answers2

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Try to write it as the product of $k$ fractions, each one is increasing.

Empy2
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$$ \binom n k /n^k > \binom {n-1}k /(n-1)^k \\\Leftrightarrow\\ \frac n {n-k}\binom {n-1} k /n^k > \binom {n-1}k /(n-1)^k \\\Leftrightarrow\\ \frac n {n-k}/n^k > 1/(n-1)^k \\\Leftrightarrow\\ \frac n {n-k} > \left (\frac n{n-1}\right)^k $$ Now fix $k$. For $n=k$ the inequality holds.

All that's left to show is that the slope of the LHS is bigger than that of the RHS once the LHS is bigger than the RHS.

Sudix
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