One Plus Reciprocal to the Nth

Theorem
Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as $x_n = \left({1 + \dfrac 1 n}\right)^n$.

Then $\left \langle {x_n} \right \rangle$ converges to a limit as $n$ increases without bound.

Proof
First we show that $\left \langle {x_n} \right \rangle$ is increasing.

Let $a_1 = a_2 = \cdots = a_{n-1} = 1 + \dfrac 1 {n-1}$ and 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:
 * $A_n = \dfrac{\left({n - 1}\right) \left({1 + \dfrac 1 {n-1}}\right) + 1} n = \dfrac {n + 1} n = 1 + \dfrac 1 n$
 * $G_n = \left({1 + \dfrac 1 {n - 1}}\right)^{\dfrac {n - 1} n}$

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

Thus:
 * $\displaystyle \left({1 + \frac 1 {n-1}}\right)^{\frac {n-1} n} \le 1 + \frac 1 n$

... and so:
 * $\displaystyle x_{n-1} = \left({1 + \frac 1 {n - 1}}\right)^{n - 1} \le \left({1 + \frac 1 n}\right)^n = x_n$

Hence $\left \langle {x_n} \right \rangle$ is increasing.

Next, we show that $\left \langle {x_n} \right \rangle$ is bounded above.

Using the Binomial Theorem:

So $\left \langle {x_n} \right \rangle$ is bounded above by $3$.

From the Monotone Convergence Theorem (Real Analysis), it follows that $\left \langle {x_n} \right \rangle$ 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).