What is wrong with this proof of $e \le 2$

Question

For every $n$, $e \le \left(1 + \frac{1}{n}\right)^n$.

$$ \left(1 + \frac{1}{n}\right)^n = 1 + \frac{n}{1!}\times\frac{1}{n} + \frac{n(n-1)}{2!}\times\frac{1}{n^2}+\ldots+ \frac{n!}{n!}\times\frac{1}{n^n} \le 1+\frac{1}{1!}+\frac{1}{2!}+\ldots+\frac{1}{n!}\le1+\frac{1}{2}+\frac{1}{2^2}+\ldots+\frac{1}{2^n} $$

As $n\rightarrow\infty$, we get $$ e \le e \le 2 $$

$2$ is the sum of the geometric series on the right side of the given inequality. And we get $e$ in the middle because $\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n=e$.

Answer 1

You simply missed a $1$ in your sum. Properly,

$1+(1/1!)+(1/2!)+(1/3!)+...<1\color{blue}{+1}+(1/2)+(1/2^2)+...$

Thus the comparison sum is $1+2=3$ instead of $2$.

Answer 2

The final inequality is false. In particular, the second sumand of the left hand side is $\frac{1}{1!}$, while the second sumand of the right hand side is $\frac12$, which is a smaller number.

Answer 3

Using this method (with a little generalization), you can obtain tight bounds on $e$.

For instance,

$$\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}<e<\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}\left(1+\frac15+\frac1{5^2}+\cdots\right)$$

$$\frac{65}{24}<e<\frac{87}{32}.$$