Rationale for validity of the binomial expansion involving rational powers

Question

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

Consider the following expression:

$$\sqrt{1-2x}$$

This can be rewritten as:

$$\left(1-2x\right)^\frac{1}{2}$$

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.

Answer 1

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.

Answer 2

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.