Derive the binomial series for square root from the generalized binomial theorem

Q: What theorem gives the series for sqrt(1+x)?

The generalized binomial theorem gives the expansion of (1+x)^alpha as a power series, and setting alpha=1/2 produces the series for sqrt(1+x).

Q: Why do odd products appear in the coefficients?

When alpha=1/2, the falling product (1/2)(1/2-1)⋯ creates half-integers. Rewriting those terms leads naturally to products like 1\cdot3\cdot5⋯ and double-factorial-style expressions.

Question

We know that:
$$
(1+x)^\alpha = 1 + \sum_{n=1}^{+\infty} \frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n
$$
We also know that:
$$
1	imes 3	imes \cdots 	imes (2n-3)=\frac{1	imes 2	imes 3	imes 4	imes \cdots 	imes (2n-3)	imes (2n-2)}{2	imes 4	imes \cdots 	imes (2n-2)}
$$
Plugging in $\alpha=\frac{1}{2}$:
$$
\sqrt{(1+x)} = 1 + \sum_{n=1}^{+\infty} \frac{\frac{1}{2}(\frac{1}{2}-1)\cdots(\frac{1}{2}-n+1)}{n!}x^n
$$
$$
= 1 + \sum_{n=1}^{+\infty} \frac{(-1)^{n-1}}{n!2^n}(1	imes 3	imes \cdots 	imes (2n-3))x^n
$$
$$
= 1 + \sum_{n=1}^{+\infty} \frac{(-1)^{n-1}}{n!2^n}\cdot \frac{(2n-1)!}{2^{n-1}(n-1)!}x^n
$$
$$
= 1 + \sum_{n=1}^{+\infty} \frac{(-1)^{n-1}(2n)!}{(n!)^2 4^n(2n-1)}x^n
$$

Acemy AI · Accepted Answer

> **Key concept**: This is a classic binomial-series derivation. The main idea is to plug in $\alpha=	frac12$ and then rewrite the coefficient so it looks more usable.

### Step by step
Start from the generalized binomial theorem:
$$
(1+x)^\alpha = 1 + \sum_{n=1}^{\infty} \frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n.
$$
Now set $\alpha=	frac12$:
$$
\sqrt{1+x}=(1+x)^{1/2}=1+\sum_{n=1}^{\infty}\frac{\frac12(\frac12-1)\cdots(\frac12-n+1)}{n!}x^n.
$$

### 1) Rewrite the product
The factors are
$$
\frac12,\; -\frac12,\; -\frac32,\; \ldots,\; \left(\frac12-n+1ight).
$$
So the numerator becomes
$$
\frac12\left(-\frac12ight)\left(-\frac32ight)\cdots\left(\frac12-n+1ight).
$$
A standard simplification gives
$$
\frac{\frac12(\frac12-1)\cdots(\frac12-n+1)}{n!}
=\frac{(-1)^{n-1}(1\cdot3\cdots(2n-3))}{2^n n!}.
$$
Thus
$$
\sqrt{1+x}=1+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2^n n!}(1\cdot3\cdots(2n-3))x^n.
$$

### 2) Convert the odd product
Use
$$
1\cdot3\cdots(2n-3)=\frac{1\cdot2\cdot3\cdots(2n-3)(2n-2)}{2\cdot4\cdot\cdots\cdot(2n-2)}.
$$
The numerator is $(2n-2)!$ and the denominator is
$$
2\cdot4\cdots(2n-2)=2^{n-1}(n-1)!.
$$
So
$$
1\cdot3\cdots(2n-3)=\frac{(2n-2)!}{2^{n-1}(n-1)!}.
$$
Substitute:
$$
\sqrt{1+x}=1+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2^n n!}\cdot\frac{(2n-2)!}{2^{n-1}(n-1)!}x^n.
$$
After simplifying the factorials into the common closed form, we get
$$
\sqrt{1+x}=1+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(2n)!}{(n!)^2 4^n(2n-1)}x^n.
$$

### 3) What this means in practice
So the coefficient of $x^n$ in $\sqrt{1+x}$ is
$$
\frac{(-1)^{n-1}(2n)!}{(n!)^2 4^n(2n-1)}.
$$
If you only need the first few terms, you can also write
$$
\sqrt{1+x}=1+\frac12x-\frac18x^2+\frac{1}{16}x^3-\frac{5}{128}x^4+\cdots.
$$

### Pitfall alert
A small but common slip is losing a minus sign in the product $\frac12(\frac12-1)\cdots$. Another one is mixing up $2^{n-1}(n-1)!$ with $2^n n!$ when rewriting the odd double factorial. Those factor changes are where most algebra errors happen.

### Try different conditions
If you substitute $\alpha=-	frac12$, the same theorem gives a series for $\frac1{\sqrt{1+x}}$. The coefficient pattern changes, but the method is identical: plug in the exponent, simplify the falling product, and then rewrite the factorial form into a standard coefficient formula.

### Further reading
generalized binomial theorem, power series expansion, double factorial

Question

Derive the binomial series for square root from the generalized binomial theorem

Expert Verified Solution

Step by step

1) Rewrite the product

2) Convert the odd product

3) What this means in practice

Pitfall alert

Try different conditions

Further reading

FAQ

What theorem gives the series for sqrt(1+x)?

Why do odd products appear in the coefficients?