Editing Binomial theorem (section)

=== Series for ''e'' ===
The [[e (mathematical constant)|number {{mvar|e}}]] is often defined by the formula
<math display="block">e = \lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n.</math>

Applying the binomial theorem to this expression yields the usual [[infinite series]] for {{mvar|e}}. In particular:
<math display="block">\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}.</math>

The {{mvar|k}}th term of this sum is
<math display="block">{n \choose k}\frac{1}{n^k} = \frac{1}{k!}\cdot\frac{n(n-1)(n-2)\cdots (n-k+1)}{n^k}</math>

As {{math|''n'' → ∞}}, the rational expression on the right approaches {{math|1}}, and therefore
<math display="block">\lim_{n\to\infty} {n \choose k}\frac{1}{n^k} = \frac{1}{k!}.</math>

This indicates that {{mvar|e}} can be written as a series:
<math display="block">e=\sum_{k=0}^\infty\frac{1}{k!}=\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots.</math>

Indeed, since each term of the binomial expansion is an [[Monotonic function|increasing function]] of {{mvar|n}}, it follows from the [[monotone convergence theorem]] for series that the sum of this infinite series is equal to&nbsp;{{mvar|e}}.