# 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:

$A_n = \dfrac{\left({n - 1}\right) \left({1 + \dfrac 1 {n-1}}\right) + 1} n = \dfrac {n + 1} n = 1 + \dfrac 1 n$
$G_n = \left({1 + \dfrac 1 {n - 1}}\right)^{\dfrac {n - 1} n}$
$G_n \le A_n$

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:

 $$\displaystyle \left({1 + \frac 1 n}\right)^n$$ $$=$$ $$\displaystyle 1 + n \left({\frac 1 n}\right) + \frac {n \left({n-1}\right)} 2 \left({\frac 1 n}\right)^2 + \cdots + \left({\frac 1 n}\right)^n$$ $\quad$ $\quad$ $$\displaystyle$$ $$=$$ $$\displaystyle 1 + 1 + \left({1 - \frac 1 n}\right) \frac 1 {2!} + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right)\frac 1 {3!} + \cdots + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right) \cdots \left({1 - \frac {n-1} n}\right) \frac 1 {n!}$$ $\quad$ $\quad$ $$\displaystyle$$ $$\le$$ $$\displaystyle 1 + 1 + \frac 1 {2!} + \frac 1 {3!} + \cdots + \frac 1 {n!}$$ $\quad$ $\quad$ $$\displaystyle$$ $$\le$$ $$\displaystyle 1 + 1 + \frac 1 2 + \frac 1 {2^2} + \cdots + \frac 1 {2^n}$$ $\quad$ (because $2^{n - 1} \le n!$) $\quad$ $$\displaystyle$$ $$=$$ $$\displaystyle 1 + \frac {1 - \left({\frac 1 2}\right)^n} {1 - \frac 1 2}$$ $\quad$ $\quad$ $$\displaystyle$$ $$=$$ $$\displaystyle 1 + 2 \left({1 - \left({\frac 1 2}\right)^n}\right)$$ $\quad$ $\quad$ $$\displaystyle$$ $$<$$ $$\displaystyle 3$$ $\quad$ $\quad$

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).