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# convergence of the sequence (1+1/n)^n

###### Theorem 1.

The following sequence:

$a_{n}=\left(1+\frac{1}{n}\right)^{n}$ | (1) |

is convergent.

###### Proof.

The proof will be given by demonstrating that the sequence (1) is:

1. monotonic (increasing), that is $a_{{n}}<a_{{n+1}}$

2. bounded above, that is $\forall n\in\mathbb{N},a_{n}<M$ for some $M>0$

In order to prove part 1, consider the binomial expansion for $a_{n}$:

$a_{n}=\sum_{{k=0}}^{{n}}\binom{n}{k}\frac{1}{n^{k}}=\sum_{{k=0}}^{{n}}\frac{1}% {k!}\frac{n}{n}\frac{n-1}{n}\ldots\frac{n-(k-1)}{n}=\sum_{{k=0}}^{{n}}\frac{1}% {k!}\left(1-\frac{1}{n}\right)\ldots\left(1-\frac{k-1}{n}\right).$ |

Since $\forall i\in\{1,2\ldots(k-1)\}:(1-\frac{i}{n})<(1-\frac{i}{n+1})$, and since the sum $a_{{n+1}}$ has one term more than $a_{n}$, it is demonstrated that the sequence (1) is monotonic.

In order to prove part 2, consider again the binomial expansion:

$a_{n}=1+\frac{n}{n}+\frac{1}{2!}\frac{n(n-1)}{n^{2}}+\frac{1}{3!}\frac{n(n-1)(% n-2)}{n^{3}}+\ldots+\frac{1}{n!}\frac{n(n-1)\ldots(n-n+1)}{n^{n}}.$ |

Since $\forall k\in\{2,3\ldots n\}:\frac{1}{k!}<\frac{1}{2^{{k-1}}}$ and $\frac{n(n-1)\ldots(n-(k-1))}{n^{k}}<1$:

$a_{n}<1+\left(1+\frac{1}{2}+\frac{1}{2\times 2}+\ldots+\frac{1}{2^{{n-1}}}% \right)<1+\left(\frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}}\right)<3-\frac{1}{2^{{% n-1}}}<3$ |

where the formula giving the sum of the geometric progression with ratio $1/2$ has been used. ∎

In conclusion, we can say that the sequence (1) is convergent and its limit corresponds to the supremum of the set $\{a_{n}\}\subset\left[2,3\right)$, denoted by $e$, that is:

$\lim_{{n\to\infty}}\left(1+\frac{1}{n}\right)^{n}=\sup_{{n\in\mathbb{N}}}\left% \{\left(1+\frac{1}{n}\right)^{n}\right\}\triangleq e,$ |

which is the definition of the Napier’s constant.

## Mathematics Subject Classification

33B99*no label found*

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