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Definiamo
:<math>s_n := \sum_{k=0}^n\frac{1}{k!}~ = 2 + \sum_{k=2}^n\frac{1}{k!}~,</math>
:<math>t_n = \left(1+\frac{1}{n}\right)^n ~.</math>
Il teoremaDal [[teorema binomiale]] dice che,
:<math>(a+b)^n = \sum_{k=0}^n {n \choose k}a^{n-k}b^{k}</math>
Ponendo
:<math>a = 1</math>
:<math>b = \frac{1}{n}</math>
:<math>t_n=\left(1+\frac{1}{n}\right)^n ~ = \sum_{k=0}^n {n \choose k}1^{n-k}n^{-k} = 2 + \sum_{k=2}^n {n \choose k}\frac{1}{n^{k}}</math> ▼
▲:<math>t_n= \left(1+\frac{1}{n}\right)^n ~ = \sum_{k=0}^n {n \choose k} \frac{1 ^}{ n-k}n^ {-k} = 2 1+1+ \sum_{k=2}^n \frac{n (n-1)(n-2)\ choose cdots(n-(k }\frac{-1 ))}{ k!\,n^ {k }}</math>
Dato che
<math>s_n = t_n</math>
:<math>2 + \sum_{k=2}^n\frac{1}{k!}~ = 2 + \sum_{k=2}^n {n \choose k}\frac{1}{n^{k}} = 2 + \sum_{k=2}^n \frac{n!}{k! \cdot \left( n - k \right)!} \frac{1}{n^{k}} = 2 + \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)\le s_n</math>
tale che
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