User:Thpigdog/Limit power identity

The identity,
 * $ \displaystyle \lim_{n \to \infty}(1 + \frac{a}{n} + \frac{b}{n^2})^n = \lim_{n \to \infty}(1 + \frac{a}{n})^n $

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

To prove start with,
 * $ \displaystyle \lim_{n \to \infty}(1 + \frac{a}{n} + \frac{b}{n^2})^n - (1 + \frac{a}{n})^n $

Firstly use this theorem. For x and y real and m > 0 then,
 * $ x < y \implies x^m < y^m $

Then as,
 * $1 + \frac{a}{n} < 1 + \frac{a}{n} + \frac{b}{n^2}$

implies,
 * $ (1 + \frac{a}{n})^m < (1+\frac{a}{n}+\frac{b}{n^2})^m $

and so,
 * $ \displaystyle 0 <= \lim_{n \to \infty}(1 + \frac{a}{n} + \frac{b}{n^2})^n - (1 + \frac{a}{n})^n $

Use the identity,
 * $ a^n - b^n = (a-b) \sum_{k=0}^{n-1} a^k b^{n-1-k} $

to get,
 * $ \displaystyle \lim_{n \to \infty}((1 + \frac{a}{n} + \frac{b}{n^2}) - (1 + \frac{a}{n})) \sum_{k=0}^{n-1} (1+\frac{a}{n}+\frac{b}{n^2})^k (1 + \frac{a}{n})^{n-1-k}$

Which simplifies to,
 * $ \displaystyle \lim_{n \to \infty}\frac{b}{n^2} \sum_{k=0}^{n-1} (1+\frac{a}{n}+\frac{b}{n^2})^k (1 + \frac{a}{n})^{n-1-k}$

Reusing,
 * $ (1 + \frac{a}{n})^m < (1+\frac{a}{n}+\frac{b}{n^2})^m $

Then
 * $ \displaystyle \lim_{n \to \infty}\frac{b}{n^2} \sum_{k=0}^{n-1} (1+\frac{a}{n}+\frac{b}{n^2})^k (1 + \frac{a}{n})^{n-1-k}$
 * $ <= \displaystyle \lim_{n \to \infty}\frac{b}{n^2} \sum_{k=0}^{n-1} (1+\frac{a}{n}+\frac{b}{n^2})^k (1 + \frac{a}{n}+\frac{b}{n^2})^{n-1-k}$


 * $ = \displaystyle \lim_{n \to \infty}\frac{b}{n^2} \sum_{k=0}^{n-1} (1+\frac{a}{n}+\frac{b}{n^2})^{n-1} $


 * $ = \displaystyle \lim_{n \to \infty}\frac{b}{n^2} n (1+\frac{a}{n}+\frac{b}{n^2})^{n-1} $


 * $ = \displaystyle \lim_{n \to \infty}\frac{b}{n} (1+\frac{a}{n}+\frac{b}{n^2})^{n-1} $


 * $ = 0$

if $\displaystyle \lim_{n \to \infty}(1+\frac{a}{n}+\frac{b}{n^2})^{n-1}$ is finite.

So,
 * $ \displaystyle 0 <= \lim_{n \to \infty}(1 + \frac{a}{n} + \frac{b}{n^2})^n - (1 + \frac{a}{n})^n <= 0$

and so,
 * $ \displaystyle \lim_{n \to \infty}(1 + \frac{a}{n} + \frac{b}{n^2})^n - (1 + \frac{a}{n})^n = 0$

To show that $\displaystyle \lim_{n \to \infty}(1+\frac{a}{n}+\frac{b}{n^2})^{n-1}$ is finite, choose some positive real c. Then in the limit as n goes to infinity.
 * $1+\frac{a}{n}+\frac{b}{n^2} < 1+\frac{a+c}{n}$

so,
 * $0 <= \displaystyle \lim_{n \to \infty}(1+\frac{a}{n}+\frac{b}{n^2})^{n-1} <= \lim_{n \to \infty}(1+\frac{a+c}{n})^n$

But by Equivalence of Definitions of Exponential Function,
 * $\displaystyle \lim_{n \to \infty}(1+\frac{a+c}{n})^n = e^{a+c}$

which is finite for finite parameters.

This proves the identity,
 * $ \displaystyle \lim_{n \to \infty}(1 + \frac{a}{n} + \frac{b}{n^2})^n = \lim_{n \to \infty}(1 + \frac{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.