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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?

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  • $\begingroup$ The expansion $$\frac1{1+u}=\sum_n(-1)^nu^n$$ upon which yours is built, is valid for $$|u|<1$$ Is this clear to you? $\endgroup$
    – Did
    Sep 2, 2017 at 14:13

2 Answers 2

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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.

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Because $\frac{1}{(1+4x)^2}={\left (\frac{1}{1+4x} \right)^2}$, and it is convergent iff $\frac{1}{1+4x} $ is absolutely convergent.

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