One Plus Reciprocal to the Nth

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

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 $$a_1 = a_2 = \cdots = a_{n-1} = 1 + \frac 1 {n-1}$$ 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:
 * $$A_n = \frac 1 n \left({n - 1}\right) \left({1 + \frac 1 {n-1}}\right) + 1 = \frac {n - 1} n = 1 + \frac 1 n$$;
 * $$G_n = \left({1 + \frac 1 {n-1}}\right)^{\frac {n-1}{n}}$$.

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

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

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

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).