How to expand $(1+ \frac{1}{x})^n$?

Question

I recently saw the expansion $(1+ \frac{1}{x})^n = 1 + \frac{n}{x} + \frac{n(n-1)}{2!x^2} + \frac{n(n-1)(n-2)(n-3)}{3!x^3}.... $ where $n \in \mathbb Q$

From what I understood, they have taken the Taylor series of $(1+x)^n$ and put $x=\frac{1}{t}$. This doesn't make sense to me because the Taylor series used, uses successive derivatives at zero but derivatives at zero won't be defined for $f(x)=\frac{1}{x}$.

How can I directly calculate the Taylor series for $(1+ \frac{1}{x})^n$?

Answer 1

This equation is not a Taylor series, but it is correct, at least for $x \geq 1$.

The usual Taylor series

$$ (1+t)^n = 1 + nt + \frac{n(n-1)}{2!} t^2 + \frac{n(n-1)(n-2)}{3!} t^3 + \cdots $$

is a true equation for every rational $n$ and every real $t$ with $-1 < t \leq 1$, in the sense that the infinite series converges to $(1+t)^n$. (Also for some other domains, but in general we may need to worry about when $a^b$ even has a clear meaning.) This fact is related to repeated derivatives, but we don't need those derivatives to just say that the equation holds true.

If $x \geq 1$, then $0 < \frac{1}{x} \leq 1$. So just by substitution, it is true that

$$ \left(1+\frac{1}{x}\right)^n = 1 + \frac{n}{x} + \frac{n(n-1)}{2!\, x^2} + \frac{n(n-1)(n-2)}{3!\, x^3} + \cdots $$

Answer 2

Expand $(x+1)^n$ as a Taylor series. Then divide through by $x^n$.