stylistic:tone
Series for e

The number e is often defined by the formula e = lim n → ∞ (1 + 1 n) n. {\displaystyle e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}.}

Applying the binomial theorem to this expression yields the usual infinite series for e. In particular: (1 + 1 n) n = 1 + (n 1) 1 n + (n 2) 1 n 2 + (n 3) 1 n 3 + ⋯ + (n n) 1 n n. {\displaystyle \left(1+{\frac {1}{n}}\right)^{n}=1+{n \choose 1}{\frac {1}{n}}+{n \choose 2}{\frac {1}{n^{2}}}+{n \choose 3}{\frac {1}{n^{3}}}+\cdots +{n \choose n}{\frac {1}{n^{n}}}.}

The k th term of this sum is (n k) 1 n k = 1 k ! ⋅ n (n − 1) (n − 2) ⋯ (n − k + 1) n k {\displaystyle {n \choose k}{\frac {1}{n^{k}}}={\frac {1}{k!}}\cdot {\frac {n(n-1)(n-2)\cdots (n-k+1)}{n^{k}}}}

As n → ∞, the rational expression on the right approaches 1, and therefore lim n → ∞ (n k) 1 n k = 1 k !. {\displaystyle \lim _{n\to \infty }{n \choose k}{\frac {1}{n^{k}}}={\frac {1}{k!}}.}

This indicates that e can be written as a series: e = ∑ k = 0 ∞ 1 k ! = 1 0 ! + 1 1 ! + 1 2 ! + 1 3 ! + ⋯. {\displaystyle e=\sum _{k=0}^{\infty }{\frac {1}{k!}}={\frac {1}{0!}}+{\frac {1}{1!}}+{\frac {1}{2!}}+{\frac {1}{3!}}+\cdots.}

Indeed, since each term of the binomial expansion is an increasing function of n, it follows from the monotone convergence theorem for series that the sum of this infinite series is equal to e.