One Plus Reciprocal to the Nth

Theorem
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = \paren {1 + \dfrac 1 n}^n$.

Then $\sequence {x_n}$ converges to a limit as $n$ increases without bound.

Proof
First we show that $\sequence {x_n}$ is increasing.

Let $a_1 = a_2 = \cdots = a_{n - 1} = 1 + \dfrac 1 {n - 1}$.

Let $a_n = 1$.

Let:
 * $A_n$ be the arithmetic mean of $a_1 \ldots a_n$
 * $G_n$ be the geometric mean of $a_1 \ldots a_n$

Thus:
 * $A_n = \dfrac {\paren {n - 1} \paren {1 + \dfrac 1 {n - 1} } + 1} n = \dfrac {n + 1} n = 1 + \dfrac 1 n$
 * $G_n = \paren {1 + \dfrac 1 {n - 1} }^{\dfrac {n - 1} n}$

By Cauchy's Mean Theorem‎:
 * $G_n \le A_n$

Thus:
 * $\paren {1 + \dfrac 1 {n - 1} }^{\frac {n - 1} n} \le 1 + \dfrac 1 n$

and so:
 * $x_{n - 1} = \paren {1 + \dfrac 1 {n - 1} }^{n - 1} \le \paren {1 + \dfrac 1 n}^n = x_n$

Hence $\sequence {x_n}$ is increasing.

Next, we show that $\sequence {x_n}$ is bounded above.

Using the Binomial Theorem:

So $\sequence {x_n}$ is bounded above by $3$.

From the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {x_n}$ converges to a limit.

Also see
Note that, although we have proved that this sequence converges to some limit less than $3$ (and incidentally greater than $2$), we have not at this stage determined exactly what this number actually is.

See Euler's number, where this sequence provides a definition of that number (one of several that are often used).