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:

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


1 Answer 1


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.