More accurate estimation of mathematical constant $e$

Question

Very often in books and also on Wikipedia we can find that: $$e \approx \left(1+\frac{1}{n}\right)^n$$ but I want more accurate estimation, it means instead using $\approx$ I wonder if I can use $\leq$, $\geq$ or $>,<$ and if it is possible how can I show that?

Answer 1

The following inequality holds: $$(1+1/n)^n < e < (1+1/n)^{n+1}.$$

Answer 2

The approximation comes from the fact, that

$$lim_{n->\infty}(1+\frac{1}{n})^n = e$$

Answer 3

The following inequalities hold : $$\frac{2n+2}{2n+1}\left(1+\frac 1n\right)^n\lt e\lt\frac{2n+1}{2n}\left(1+\frac 1n\right)^n,$$ $$\frac{2n}{2n+1}\left(1+\frac 1n\right)^{n+1}\lt e\lt\frac{2n+1}{2n+2}\left(1+\frac 1n\right)^{n+1}.$$

Also, $$e\lt\left(1+\frac 1n\right)^{n+(1/2)}\lt e\times\exp\frac{1}{4n(n+1)}.$$