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

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Theorem

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

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

is convergent.


Proof

From Cauchy's Mean Theorem:

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

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

Setting:

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

and:

$a_n = 1$

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

\(\displaystyle \paren {1 + \dfrac x {n - 1} }^{\frac {n - 1} n}\) \(\le\) \(\displaystyle \dfrac {\paren {n - 1} \paren {1 + \frac x {n - 1} } + 1} n\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + \dfrac x n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {1 + \dfrac x {n - 1} }^{n - 1}\) \(\le\) \(\displaystyle \paren {1 + \dfrac x n}^n\)

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:

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

that is, when:

$n \ge 1 - y$

From Equivalence of Definitions of Real Exponential Function: Limit of Sequence implies Sum of Series, we have:

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

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:

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

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.

$\blacksquare$


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