convergence of the sequence (1+1/n)^n
Theorem 1.
Proof.
The proof will be given by demonstrating that the sequence (1) is:

1.
monotonic (increasing), that is $$

2.
bounded above, that is $$ for some $M>0$
In order to prove part 1, consider the binomial expansion for ${a}_{n}$:
$${a}_{n}=\sum _{k=0}^{n}\left(\genfrac{}{}{0pt}{}{n}{k}\right)\frac{1}{{n}^{k}}=\sum _{k=0}^{n}\frac{1}{k!}\frac{n}{n}\frac{n1}{n}\mathrm{\dots}\frac{n(k1)}{n}=\sum _{k=0}^{n}\frac{1}{k!}\left(1\frac{1}{n}\right)\mathrm{\dots}\left(1\frac{k1}{n}\right).$$ 
Since $$, 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:
$${a}_{n}=1+\frac{n}{n}+\frac{1}{2!}\frac{n(n1)}{{n}^{2}}+\frac{1}{3!}\frac{n(n1)(n2)}{{n}^{3}}+\mathrm{\dots}+\frac{1}{n!}\frac{n(n1)\mathrm{\dots}(nn+1)}{{n}^{n}}.$$ 
Since $$ and $$:
$$ 
where the formula^{} giving the sum of the geometric progression with ratio $1/2$ has been used. ∎
In conclusion^{}, we can say that the sequence (1) is convergent and its limit corresponds to the supremum of the set $\{{a}_{n}\}\subset [2,3)$, denoted by $e$, that is:
$$\underset{n\to \mathrm{\infty}}{lim}{\left(1+\frac{1}{n}\right)}^{n}=\underset{n\in \mathbb{N}}{sup}\left\{{\left(1+\frac{1}{n}\right)}^{n}\right\}\triangleq e,$$ 
which is the definition of the Napier’s constant.
Title  convergence of the sequence (1+1/n)^n 

Canonical name  ConvergenceOfTheSequence11nn 
Date of creation  20130322 17:43:26 
Last modified on  20130322 17:43:26 
Owner  kfgauss70 (18761) 
Last modified by  kfgauss70 (18761) 
Numerical id  7 
Author  kfgauss70 (18761) 
Entry type  Theorem 
Classification  msc 33B99 
Related topic  NondecreasingSequenceWithUpperBound 