User:Thpigdog/Limit power identity

The identity:

$\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac a n + \dfrac b {n^2} }^n = \lim_{n \mathop \to \infty} \paren {1 + \frac a n}^n$

for $a$, $b$ real and $b > 0$.

To prove start with,

$\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac a n + \dfrac b {n^2} }^n - \paren {1 + \dfrac a n}^n$

Firstly use this theorem. For $x$ and $y$ real and $m > 0$:

$x < y \implies x^m < y^m$

Then, as:

$1 + \dfrac a n < 1 + \dfrac a n + \dfrac b {n^2}$

implies:

$\paren {1 + \frac a n}^m < \paren {1 + \dfrac a n + \dfrac b {n^2} }^m$

and so:

$\ds 0 \le \lim_{n \mathop \to \infty} \paren {1 + \dfrac a n + \frac b {n^2} }^n - \paren {1 + \dfrac a n}^n$

Use the identity,

$\ds a^n - b^n = \paren {a - b} \sum_{k \mathop = 0}^{n - 1} a^k b^{n - 1 - k}$

to get:

$\ds \lim_{n \mathop \to \infty} \paren {\paren {1 + \dfrac a n + \dfrac b {n^2} } - \paren {1 + \dfrac a n} } \sum_{k \mathop = 0}^{n - 1} \paren {1 + \dfrac a n + \dfrac b {n^2} }^k \paren {1 + \dfrac a n}^{n - 1 - k}$

which simplifies to:

$\ds \lim_{n \mathop \to \infty} \dfrac b {n^2} \sum_{k \mathop = 0}^{n - 1} \paren {1 + \dfrac a n + \dfrac b {n^2} }^k \paren {1 + \dfrac a n}^{n - 1 - k}$

Reusing:

$\paren {1 + \dfrac a n}^m < \paren {1 + \dfrac a n + \dfrac b {n^2} }^m$

Then:

$\ds \lim_{n \mathop \to \infty} \dfrac b {n^2} \sum_{k \mathop = 0}^{n - 1} \paren {1 + \dfrac a n + \dfrac b {n^2} }^k \paren {1 + \dfrac a n}^{n - 1 - k}$

$\le \ds \lim_{n \mathop \to \infty} \dfrac b {n^2} \sum_{k\mathop = 0}^{n - 1} \paren {1 + \dfrac a n + \dfrac b {n^2} }^k \paren {1 + \dfrac a n + \dfrac b {n^2} }^{n - 1 - k}$

$= \ds \lim_{n \mathop \to \infty} \dfrac b {n^2} \sum_{k \mathop = 0}^{n - 1} \paren {1 + \dfrac a n + \dfrac b {n^2} }^{n - 1}$

$= \ds \lim_{n \mathop \to \infty} \dfrac b {n^2} n \paren {1 + \dfrac a n + \dfrac b {n^2} }^{n - 1}$

$= \ds \lim_{n \mathop \to \infty} \dfrac b n \paren {1 + \dfrac a n + \dfrac b {n^2} }^{n - 1}$

$= 0$

if $\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac a n + \dfrac b {n^2} }^{n - 1}$ is finite.

So:

$\ds 0 \le \lim_{n \mathop \to \infty} \paren {1 + \dfrac a n + \dfrac b {n^2} }^n - \paren {1 + \dfrac a n}^n \le 0$

and so:

$\ds \lim_{n \mathop \to \infty}\paren {1 + \frac a n + \frac b {n^2} }^n - \paren {1 + \frac a n}^n = 0$

To show that $\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac a n + \dfrac b {n^2} }^{n - 1}$ is finite, choose some positive real $c$. Then in the limit as $n$ goes to infinity:

$1 + \dfrac a n + \dfrac b {n^2} < 1 + \dfrac {a + c} n$

so:

$0 \le \ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac a n + \dfrac b {n^2} }^{n - 1} \le \lim_{n \mathop \to \infty} \paren {1 + \dfrac {a + c} n}^n$

But by Equivalence of Definitions of Exponential Function,

$\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac {a + c} n}^n = e^{a + c}$

which is finite for finite parameters.

This proves the identity:

$\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac a n + \dfrac b {n^2} }^n = \lim_{n \mathop \to \infty} \paren {1 + \dfrac a n}^n$

where $a$ is real and $b$ is positive real. With the identity proven for positive real, and because both sides are analytic it must hold for all complex $a$ and $b$.