I want to ask a question about the validity of the binomial expansion involving rational powers.

Consider the following expression:


This can be rewritten as:


Now, as I can see from this polynomial, it is clear that for a convergent series to exist, we need to satisfy the validity condition:

$$\left|2x\right| < 1$$

and as a result, the validity condition that we write can be expressed as:

$$\left|x\right| < \frac{1}{2}$$

From this, I can also conclude that if

$$\left|x\right| \geq \frac{1}{2}$$

that the resulting expansion would be divergent.

However, I'm confused as to why the validity of the binomial expansion requires a convergent series to be generated. I cannot see the justification why a validity needs to specified such that the resulting series is convergent for binomial expansions such as those involving rational powers above.

No textbook or video that I've watched on the internet seems to justify this point with any clarity.

Why does the validity of a binomial expansion involving rational powers need to satisfy a convergent series as opposed to a divergent series, if convergence is (I'm assuming) the condition we are aiming to satisfy in the first place?

EDIT The function was meant to be $\sqrt{1-2x}$, not $\sqrt{1+2x}$ as previously stated.

  • $\begingroup$ What do you mean by validity? For the equation between values $(1+2x)^{1/2}=\sum_{n=0}^{\infty}\binom{1/2}{n}2^nx^2$ you need, in particular, that the right hand side represents a value, that the series converges. However, there are other contexts (local ring of formal power series, for example) in which the equation is valid. However, in that case the equation is not referring to values, but to the algebraic objects themselves. $\endgroup$
    – plop
    Aug 23, 2022 at 10:55
  • $\begingroup$ @user85667 This is my question in mind. If you watch the video in question (where the same function appears), the narrator also specifies the condition for validity to be the same as the inequality containing x to yield a convergent series. I failed to find an appropriate justification for this and hence thought to post it here to see if an answer could be found! $\endgroup$
    – vik1245
    Aug 23, 2022 at 10:58
  • $\begingroup$ See radius of convergence. They just computed it, for example, by using the formula with the ratio of two consecutive coefficients (ratio test) that you see in the link. That computation gives convergence for $|x|<2$ and divergence for $|x|>2$. It doesn't however, give divergence for $|x|=2$. $\endgroup$
    – plop
    Aug 23, 2022 at 11:01
  • $\begingroup$ @user85667 why can't the expansion be "valid" when $\left|x\right| > 2$? This is the question I'm struggling to answer at the moment and is at the crux of the question itself. I see that if $\left|x\right| > 2$ then divergence exists although I find that not to be a sufficient justification in itself (I hope that makes sufficient sense). $\endgroup$
    – vik1245
    Aug 23, 2022 at 11:09
  • $\begingroup$ As an equation between values, if you plug in an $x$ such that $|x|>2$, the left hand side is a real number, while the right hand side (using the most common definition for sum of a series) is not a real number, it is not defined. For example, for $x=-4$, the right hand side would be infinite, while the left could be $\sqrt{7}i$. $\endgroup$
    – plop
    Aug 23, 2022 at 11:11

2 Answers 2


This is one way of approaching the question at hand which I feel may be more accessible to the layman reader.

I realised that one application of the binomial expansion is to approximate the value of a function for a given value $x$. This is implicitly discussed in the video above but not explicitly stated.

For this to be possible, successive terms should not cause the value of the sum to diverge from a given result but instead converge towards a specific value e.g. $x^3$, $x^4$, $x^5$ terms etc.

It is for this reason that a convergent series is desired, and as a result, the approximation of a function via Binomial expansion only is satisfied when an inequality in x is provided.


For $-\frac12 < x< \frac12$, $$ (1+2x)^{1/2} = 1+x-\frac{1}{2}x^2+\frac{1}{2}x^3-\frac{5}{8}x^4+\dots $$ For $x>\frac12$, $$ (1+2x)^{1/2}=2^{1/2}x^{1/2} + \frac{1}{2^{3/2}}x^{-1/2} -\frac{1}{2^{9/2}}x^{-3/2}+\frac{1}{2^{13/2}}x^{-5/2}- \frac{5}{2^{21/2}}x^{-7/2}+\dots $$ Both are "convergent" series.

  • $\begingroup$ As I expressed in a previous comment, I do apologise – the function should have been $\sqrt{1-2x}$ – does this change the answer at all in any way? $\endgroup$
    – vik1245
    Aug 23, 2022 at 12:03
  • $\begingroup$ $\sqrt{1-2x}$ is defined (as a real number) only for $x<\frac12$. So of course the series does not extend beyond $\frac12$. We could get something similar to this answer valid for $x<-\frac12$. $\endgroup$
    – GEdgar
    Aug 23, 2022 at 14:20

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