二項定理
ネイピア数 e を極限
e = lim n → ∞ ( 1 + 1 n ) n {\displaystyle e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}}
で定義するとき、二項定理と単調収束定理を用いれば e の級数表示を得る。
( 1 + 1 n ) n = 1 + ( n 1 ) 1 n + ( n 2 ) 1 n 2 + ( n 3 ) 1 n 3 + ⋯ + ( n n ) 1 n n = 1 + 1 + 1 2 ! ( 1 − 1 n ) + 1 3 ! ( 1 − 1 n ) ( 1 − 2 n ) + ⋯ + 1 n ! ( 1 − 1 n ) ( 1 − 2 n ) ⋯ ( 1 − n − 1 n ) {\displaystyle {\begin{aligned}\left(1+{\frac {1}{n}}\right)^{n}&=1+{\binom {n}{1}}{\frac {1}{n}}+{\binom {n}{2}}{\frac {1}{n^{2}}}+{\binom {n}{3}}{\frac {1}{n^{3}}}+\cdots +{\binom {n}{n}}{\frac {1}{n^{n}}}\\&=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)\left(1-{\frac {2}{n}}\right)\cdots \left(1-{\frac {n-1}{n}}\right)\end{aligned}}}