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.

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:
 * $\displaystyle A_n = \frac{\left({n - 1}\right) \left({1 + \frac 1 {n-1}}\right) + 1} n = \frac {n + 1} n = 1 + \frac 1 n$
 * $\displaystyle G_n = \left({1 + \frac 1 {n-1}}\right)^{\frac {n-1}{n}}$

By Arithmetic Mean Never Less than Geometric Mean‎, $G_n \le A_n$.

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.

We use the Binomial Theorem:

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


 * So, from the Monotone Convergence Theorem, $\left \langle {x_n} \right \rangle$ converges to a limit.

Note
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).