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I would like to prove that for $q$ any real number with $0<q<1$ and $n$ a natural number $n\geq 2$ we have:

$$\left(1-\frac{1-q}{n}\right)^n > q$$

Mathematica says yes for all $n$ I have checked (up to $n=100$) and looking at the graph for some $n$ also supports it.

I tried an induction over $n$ but can't get anywhere. Note that according to Wolfram Alpha

$$\lim_{n \rightarrow \infty} \left(1-\frac{1-q}{n}\right)^n = e^{q-1}$$

There is also a series expansion on Wolfram Alpha:

$$\left(1-\frac{1-q}{n}\right)^n = \sum_{k=0}^\infty \left(\frac{-1+q}{n}\right)^k \binom{n}{k}$$

for $|\frac{1-q}{n}|<1$ which is the case here. I am not sure if any of this helps but maybe this other series expansion:

$$\left(1-\frac{1-q}{n}\right)^n = \sum_{k=0}^\infty \frac{n^k \log^k\left(1-\frac{1-q}{n}\right)}{k!}$$

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Brenoulli's inequality states that $$ (1+x)^n\ge 1+nx$$ for $n\in\Bbb N$ and $x\ge -1$, and we have "$>$" if $n\ge2$ and $x\ne0$.

Let $x=-\frac{1-q}{n}$.

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  • $\begingroup$ Awesome! So fast. You will be thanked in my PhD thesis now (if you don't object). $\endgroup$
    – mab
    Oct 18, 2016 at 15:36

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