An example of a Maclaurin series is the Binomial theorem for negative / fractional indices:
\[(1 + x)^{n} = 1 + nx + \frac{n(n-1)}{2!} x^{2} + \frac{n(n-1)(n-2)}{3!} x^{3} + \cdots \text{ for}\hspace{0.5em} |x| < 1\]
To see this, let us work through all the steps of calculating the Maclaurin's series
\[
f(x) = \sum_{n=0}^\infty c_n x^n
\]
for the function
\[
f(x) = (1+x)^n,
\]
where \(n \in \mathbb{R} - \mathbb{N}\).
We start with
\[
f(0) = (1+0)^n = 1.
\]
So,
\[
c_0 = f(0) = 1.
\]
Differentiating \(f\) with respect to \(x\):
\[
f^{(1)}(x) = n (1+x)^{n-1}.
\]
So,
\[
f^{(1)}(0) = n(1+0)^{n-1} = n.
\]
Thus,
\[
c_1 = f^{(1)}(0) = n.
\]
Proceeding with further differentiation:
\[
f^{(2)}(x) = n(n-1)(1+x)^{n-2}.
\]
Thus,
\[
f^{(2)}(0) = n(n-1).
\]
So,
\[
c_2 = \frac{f^{(2)}(0)}{2!} = \frac{n(n-1)}{2}.
\]
By induction, one has
\[
f^{(k)}(0) = n(n-1)(n-2) \cdots (n-k+1).
\]
Thus,
\[
c_k = \frac{n(n-1)(n-2)\cdots (n-k+1)}{k!}.
\]
Thus, we have that
\[
(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \cdots + \frac{n(n-1)\cdots (n-k+1)}{k!} x^k + \cdots
\]
The radius of convergence \(R\) can be computed by using the following formula:
\[
R = \lim_{n \to \infty} \frac{c_n}{c_{n+1}}.
\]
In this case when \(c_k = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}\), and so
\[
R = \lim_{k \to \infty} \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} \div
\frac{n(n-1)(n-2)\cdots(n-k)}{(k+1)!} = 1.
\]
Therefore,
\[
(1+x)^n = 1+ nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots + \frac{n(n-1)(n-2)\cdots (n-k+1)}{k!} x^k + \cdots
\]
holds whenever \(-1 < x < 1\).
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