# Real Sequence (1 + x over n)^n is Convergent

## Theorem

The sequence $\sequence {s_n}$ defined as:

$s_n = \paren {1 + \dfrac x n}^n$

is convergent.

## Proof

$(1): \quad \ds \paren {\prod_{k \mathop = 1}^n a_k}^{1/n} \le \frac 1 n \paren {\sum_{k \mathop = 1}^n a_k}$

for $r_1, r_2, \ldots, r_n$.

Setting:

$a_1 = a_2 = \ldots = a_{n - 1} := 1 + \dfrac x {n - 1}$

and:

$a_n = 1$

Substituting for $a_1, a_2, \ldots, a_n$ into $(1)$ gives:

 $\ds \paren {1 + \dfrac x {n - 1} }^{\frac {n - 1} n}$ $\le$ $\ds \dfrac {\paren {n - 1} \paren {1 + \frac x {n - 1} } + 1} n$ $\ds$ $=$ $\ds 1 + \dfrac x n$ $\ds \leadsto \ \$ $\ds \paren {1 + \dfrac x {n - 1} }^{n - 1}$ $\le$ $\ds \paren {1 + \dfrac x n}^n$

The above is valid only if $a_1, a_2, \ldots, a_n$ are positive.

Hence we have shown that $\sequence {s_n}$ is increasing when:

$1 + \dfrac x {n - 1} \ge 0$

that is, when:

$n \ge 1 - x$
$(2): \quad \paren {1 + \dfrac x n}^n \le 1 + \size x + \dfrac {\size x^2} {2!} + \dotsb + \dfrac {\size x^n} {n!}$

Since there exists $N$ such that:

$\forall n > N: \dfrac {\size x^n} {n!} \le \paren {\dfrac 1 2}^n$

it follows from $(2)$ that:

 $\ds \paren {1 + \dfrac x n}^n$ $\le$ $\ds 1 + \size x + \dfrac {\size x^2} {2!} + \dotsb + \dfrac {\size x^N} {N!} + \paren {\dfrac 1 2}^{N + 1} + \dotsb + \paren {\dfrac 1 2}^n$ $\ds$ $\le$ $\ds C + \dfrac {1 + \paren {\frac 1 2}^{n + 1} } {1 + \frac 1 2}$ where $C$ is some constant $\ds$ $<$ $\ds C + 2$

Hence we have that $\sequence {s_n}$ is strictly increasing and bounded above.

So by the Monotone Convergence Theorem (Real Analysis), $\sequence {s_n}$ is convergent.

As $1 + \dfrac x n$ is positive when $n$ is large enough, it follows that the limit of $\sequence {s_n}$ is positive.

$\blacksquare$