Characterizations of the exponential function

The mathematical constant e can be defined in many ways. The following three definitions are most commonly used. This article discusses why each definition makes sense, and why the definitions are independent and equivalent to each other.

Three definitions

The following proof demonstrates the equivalence of the three definitions given for e above.

Define

$s_{n}=\sum _{k=0}^{n}{\frac {1}{k!}},\ t_{n}=\left(1+{\frac {1}{n}}\right)^{n}$

By the binomial theorem,

$t_{n}=\sum _{k=0}^{n}{n \choose k}{\frac {1}{n^{k}}}=1+1+\sum _{k=2}^{n}{\frac {n(n-1)(n-2)\cdots (n-(k-1))}{k!\,n^{k}}}$

$=1+1+{\frac {1}{2!}}\left(1-{\frac {1}{n}}\right)+{\frac {1}{3!}}\left(1-{\frac {1}{n}}\right)\left(1-{\frac {2}{n}}\right)+\cdots +{\frac {1}{n!}}\left(1-{\frac {1}{n}}\right)\cdots \left(1-{\frac {n-1}{n}}\right)\leq s_{n}$

so that

$\limsup _{n\to \infty }t_{n}\leq \limsup _{n\to \infty }s_{n}=e$

Here, we must use limsup's, because we don't yet know that t_n actually converges. Now, for the other direction, note that by the above expression of t_n, if 2 ≤ m ≤ n, we have

$1+1+{\frac {1}{2!}}\left(1-{\frac {1}{n}}\right)+\cdots +{\frac {1}{m!}}\left(1-{\frac {1}{n}}\right)\left(1-{\frac {2}{n}}\right)\cdots \left(1-{\frac {m-1}{n}}\right)\leq t_{n}$

Fix m, and let n approach infinity. We get

$s_{m}=1+1+{\frac {1}{2!}}+\cdots +{\frac {1}{m!}}\leq \liminf _{n\to \infty }t_{n}$

(again, we must use liminf's because we don't yet know that t_n converges). Now, take the above inequality, let m approach infinity, and put it together with the other inequality. This becomes

$\limsup _{n\to \infty }t_{n}\leq e\leq \liminf _{n\to \infty }t_{n}\leq \limsup _{n\to \infty }t_{n}$

This completes the proof. Q.E.D.