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Let $n$ and $1\leq k \leq n$ be natural numbers. Prove the inequality $$\sum_{i=k}^n \binom{n}{i}\bigg(\frac{k}{n+1}\bigg)^i\bigg(1-\frac{k}{n+1}\bigg)^{n-i} \leq 1 - \frac{1}{e} $$

Equivalently, if $X\sim$ Bin($n$,$\frac{k}{n+1}$), prove that $\mathbb{P}[X\geq k] \leq 1 - \frac{1}{e}$.

My attempt: It may be helpful to show that the LHS tends to a limit less than $1-\frac{1}{e}$ as $n \to \infty$ (already did that) and that the LHS is an increasing function on $n$ (have not done that).


marked as duplicate by Mike Earnest, Isaac Browne, Chris Custer, NCh, JonMark Perry Apr 14 '18 at 2:33

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    $\begingroup$ In fact, it's exactly the same question, asked by the same OP. What gives? $\endgroup$ – saulspatz Apr 13 '18 at 22:28

Not a complete answer, but here are few hints:

1) $\binom{n}{k} \leq \frac{n^k}{k!}$

2) $\sum_{k=0}^{n}\frac{z^k}{k!} < e^z \ \text{if} \ z>0$

3) $\sum_{i=k}^{n} p(k) = 1 - \sum_{i=0}^{k-1}p(k)$

Hope this helps


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