# An Euler’s number binom-rational inequality.

Combining the Binomial Theorem with the sequence
$e(n) := (1+\frac{1}{n})^n$, we get the following rational inequality…

Claim. For $n> 1$ it holds that $2 < (1+1/n)^{n} < \frac{67}{2^3\cdot 3} = 2.7916666666666666666\cdots$
Proof.
Applying the binomial theorem,
$\displaystyle \left(1+\frac{1}{n}\right)^n = \sum_{k=0}^{n}\binom{n}{k}\frac{1}{n^k} =$ $\displaystyle = 1+\sum_{k=1}^{n}\binom{n}{k}\frac{1}{n^k}$
$\displaystyle = 1 + \sum_{k=1}^{n} \frac{1}{k!} \frac{(n-k+1)\cdots (n-1)n}{n^k}=1+\sum_{k=1}^{n} \frac{1}{k!} \prod_{i=0}^{k-1} \frac{n-i}{n} =$
$\displaystyle = 1+\sum_{k=1}^{n} \frac{1}{k!} \prod_{i=0}^{k-1} (1 - \frac{i}{n})$

Look at the factors of the product above: fix $i$ and define $f(n):=1-\frac{i}{n}, n\geq 1$, clearly $f(n+1)>f(n)$ for any $n \geq 1$, so $(1+\frac{1}{n})^n$ is a monotonically increasing function of $n$. This imply $\left(1+\frac{1}{n}\right)^n > 2$ provided that $n>1$.

Also $1-\frac{i}{n} < 1$, thus $1+\sum_{k=1}^{n} \frac{1}{k!} \prod_{i=0}^{k-1} (1 - \frac{i}{n}) < 1+\sum_{k=1}^{n} \frac{1}{k!}.$

Now for every $n\geq 4$ it holds $n! > 2^n$, this imply
$\displaystyle 1+\sum_{k=1}^{n} \frac{1}{k!} < 1 + (1+1/2 + 1/6) + \sum_{k=4}^{n}\frac{1}{2^k} =$
$= 1+(1+1/2+1/6)- (1+1/2+1/4+1/8) + \frac{1-1/2^{n+1}}{1-1/2} <$
$< 1+(1+1/2+1/6)- (1+1/2+1/4+1/8) + \frac{1}{1-1/2} =$
$3+1/6-1/4-1/8 = \frac{67}{2^3\cdot 3} = 2.7916666666666666666\cdots$.
Q.E.D.