Binomial Expansion conditions for valid expansion $\frac{1}{(1+4x)^2}$ I was studying Binomial expansions today and I had a question about the conditions for which it is valid.
$$\frac{1}{(1+4x)^2}$$
I was asked to find the binomial expansion, up to and including the term in $x^3$. 
Now, this is how I did the expansion. 
$$\frac{1}{(1+4x)^2}$$ 
$$=(1+4x)^{-2}$$
$$ = 1 + (-2)(4x) + \frac{(-2)(-3)}{2}16x^2 + \frac{(-2)(-3)(-4)}{6}64x^3 + ...$$
$$ = 1 -8x + 48x^2 -256x^3 + ...$$
Now, the text I am reading says 
Expansion is valid as long as $|4x| < 1 ⇒ |x| < \frac{1}{4}$
I'm confused. 
How did the text come to this conclusion?


If you look at the term in $x^n$ you will find that it is $(n+1)\cdot (-4x)^n$. Therefore if $|x|\ge \frac 14$ the terms will be increasing in absolute value, and therefore the sum will not converge.
The easy way to see that $\frac 14$ is the critical value here is to note that $x=-\frac 14$ makes the denominator of the original fraction zero, so there is no prospect of a convergent series.
