A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?

The inequality $$ e_n:=\left(1+\frac1n\right)^n\leq3-\frac1n, $$ where $n\in\mathbb{N}_+$, is certainly true, because we know, how LHS is connected with $e$. The other argument is the standard proof of boundedness of $(e_n)$, which uses the binomial theorem.

Are there any more elementary proofs of this inequality?

Solution 1:

You can use induction.

For $n=1$, clearly $2 \leq 2$. Assume it holds for some $n-1 \in \mathbb{N}$. Then,

$$\left(1+ \frac{1}{n} \right)^n = \left( 1 + \frac{1}{n} \right)^{n-1} \left( 1 + \frac{1}{n} \right) \leq \left( 1 + \frac{1}{n-1} \right)^{n-1} \left( 1 + \frac{1}{n} \right) \leq \left( 3- \frac{1}{n-1} \right) \left(1 - \frac{1}{n} \right) = 3 - \frac{3}{n} - \frac{1}{n-1} + \frac{1}{n(n-1)}.$$

It is left to show

$$-\frac{3}{n} - \frac{1}{n-1} + \frac{1}{n(n-1)} \leq -\frac{1}{n}$$

or equivalently

$$ -3n+3-n+1 \leq -n+1,$$

i.e. $n \geq 1$. Since $n-1 \in \mathbb{N}$, we have $n \geq 2$ by choice, so the inequality holds.

Solution 2:

Since $f(t)=\frac{1}{t}$ is a convex function on $\mathbb{R}^+$, we have: $$\log\left(1+\frac{1}{n}\right)=\int_{n}^{n+1}\frac{dt}{t}\leq\frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)\tag{1}$$ hence: $$ \left(1+\frac{1}{n}\right)^n \leq \exp\left(1-\frac{1}{2n+2}\right)\leq\frac{e}{1+\frac{1}{2n+2}}=\frac{2n+2}{2n+3}e \tag{2}$$ that is stronger than $ \left(1+\frac{1}{n}\right)^n \leq 3-\frac{1}{n}$ for any $n\geq 2$.

Solution 3:

We can expand using Binomial Theorem:

$\displaystyle \left(1+\frac{1}{n}\right)^n = \sum\limits_{k=0}^{n} \binom{n}{k}\frac{1}{n^k} = \sum\limits_{k=0}^{n} \frac{\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)}{k!} \le \sum\limits_{k=0}^{n} \frac{1}{k!}$

Now, $\displaystyle \sum\limits_{k=0}^{n} \frac{1}{k!} \le 2 + \sum\limits_{k=2}^{n} \frac{1}{k(k-1)} = 3 - \frac{1}{n}$

which proves the required inequality.