Binomial expansion infinite series: why need form $a^n(1+\frac{b}{a}x)^n$? I suppose my question is about limits and infinities. I have been considering the expansion of $(a+bx)^n$ for negative and fractional $n$, and my text book says to always factorise out the $a^n$ like so:

$(a+bx)^n =a^n(1+\frac{b}{a}x)^n= a^n[1+n\frac{b}{a}x+\frac{n(n-1)}{2!}(\frac{b}{a}x)^2+...]$

And then the expansion is valid for $|\frac{bx}{a}|&lt;1$

My question is, why is the $a$ term involved in thevalidity requirement? Why is it incorrect to simply expand the binomial as

$(a+bx)^n = a^n+a^{n-1}n(bx)+\frac{a^{n-2}n(n-1)}{2!}(bx)^2+...]$

and say it is valid for $|bx|&lt;1$?
My question is, why is the $a$ term involved in the validity requirement?


The requirement for convergence of this series comes from the ratio test. The absolute value of the ratio of consecutive terms of your series is $$\left|\frac{\binom{n}{k+1}\color{red}{a^{n-k-1}}(bx)^{k+1}}{\binom{n}{k}\color{red}{a^{n-k}}(bx)^k}\right|,$$ or more simply $$\left|\frac{n-k}{1+k}\color{red}{a^{-1}}bx\right|.$$ Take note of the coloured parts. That shows you why $a$ does not disappear from the condition.
This is first of all a matter of book-keeping. The $(1+bx/a)^n$ is
considered a geometric series 
$$
(1+y)^n=\sum_{i\ge 0} \binom{n}{i}y^i
$$
where $y=bx/a$ is the free variable, and where
the radius of convergence is known to be $|y|&lt;1$. So instead of
remembering a lot of different subcases of how to approach this, 
only a single criterion (parameter) comes into play.