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

2. bounded above, that is for some $M>0$

In order to prove part 1, consider the binomial expansion for $a_n$:

$$a_n=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)\frac{1}{n^k}=\sum _{k=0}^n\frac{1}{k!}\frac{n}{n}\frac{n-1}{n}\mathrm{\dots }\frac{n-(k-1)}{n}=\sum _{k=0}^n\frac{1}{k!}\left(1-\frac{1}{n}\right)\mathrm{\dots }\left(1-\frac{k-1}{n}\right).$$

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:

$$a_n=1+\frac{n}{n}+\frac{1}{2!}\frac{n(n-1)}{n^2}+\frac{1}{3!}\frac{n(n-1)(n-2)}{n^3}+\mathrm{\dots }+\frac{1}{n!}\frac{n(n-1)\mathrm{\dots }(n-n+1)}{n^n}.$$

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	2013-03-22 17:43:26
Last modified on	2013-03-22 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