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.

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