An example of a Maclaurin series is the Binomial theorem for negative / fractional indices:
\[(1 + x)^{n} = 1 + nx + \frac{n(n-1)}{2!} x^{2} + \frac{n(n-1)(n-2)}{3!} x^{3} + \cdots \text{ for}\hspace{0.5em} |x| &lt; 1\]
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&nbsp;$n \in \mathbb{R} - \mathbb{N}$.
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&nbsp;$c_k = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$, 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 $-1 &lt; x &lt; 1$.

An example of a Maclaurin series is the Binomial theorem for negative / fractional indices:

\[(1 + x)^{n} = 1 + nx + \frac{n(n-1)}{2!} x^{2} + \frac{n(n-1)(n-2)}{3!} x^{3} + \cdots \text{ for } |x| < 1\]

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 $-1 < x < 1$.

Binomial series as an example of Maclaurin's series

Binomial series as an example of Maclaurin's series