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

###### Proof.

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

1. 1.

monotonic (increasing), that is $a_{n}

2. 2.

bounded above, that is $\forall n\in\mathbb{N},a_{n} 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.

Title convergence of the sequence (1+1/n)^n ConvergenceOfTheSequence11nn 2013-03-22 17:43:26 2013-03-22 17:43:26 kfgauss70 (18761) kfgauss70 (18761) 7 kfgauss70 (18761) Theorem msc 33B99 NondecreasingSequenceWithUpperBound