Let \(f(x)=(1+x)^r\) where \(r\) is any real number. The Maclaurin series of \(f(x)\) is:
\[ 1 + rx + \frac{r(r-1)}{2!} x^2 + \frac{r(r-1)(r-2)}{3!} x^3 + \cdots+ \frac{r(r-1) \cdots (r-(n-1))}{n!}x^n + \cdots \]
Let's use the ratio test to see if this series converges:
\[\begin{align} \lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} &= \lim_{n \to \infty} \left| \frac{r(r-1) \cdots (r-n) x^{n+1}}{(n+1)!} * \frac{n!}{r(r-1) \cdots (r-(n-1)) x^n} \right| \\ &= \lim_{n \to \infty} \left| \frac{r-n}{n+1} \right| |x| \end{align}\]
Since \( \left[ \lim_{n \to \infty} \left| \frac{r-n}{n+1} \right| = 1 \right] \):
\[ \lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} = |x| \]
Which means the Maclaurin series converges if \(|x| < 1\) and diverges if \(|x| > 1\). There exists some \(c\) between 0 and \(x\), such that:
\[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \]
Let's try to find \(f^{(n+1)}(c)\):
\[\begin{align} f^{(n+1)}(x) &= r(r-1)(r-2)(r-3) \cdots (r-n)(1+x)^{r-(n+1)} \\ f^{(n+1)}(c) &= r(r-1)(r-2)(r-3) \cdots (r-n)(1+c)^{r-(n+1)} \end{align}\]
This means:
\[ R_n(x) = \frac{r(r-1)(r-2)(r-3) \cdots (r-n)(1+c)^{r-(n+1)}}{(n+1)!} x^{n+1} \]
We can rewrite this as:
\[ R_n(x) = \left( \frac{r}{n+1} \right) \left( \frac{(r-1)(r-2)(r-3) \cdots (r-n)}{(1)(2)(3)\cdots(n)} \right) \left( \frac{(1+c)^r}{(1+c)^{(n+1)}}\right) \left( x^{n+1} \right) \]
If we take the modulus of both sides:
\[ | R_n(x) | =\left| \frac{r}{n+1} \right| \left| \frac{(1-r)(2-r)(3-r) \cdots (n-r)}{(1)(2)(3)\cdots(n)} \right| \ \frac{| (1+c)^r |}{|1+c|^{(n+1)}} \ \left| x^{n+1} \right| \]