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.