# Maclaurin series is the geometric series, question

I have a question on the geometric series being represented by the Maclaurin series.

Wiki defines the Taylor series as: $$\sum_{n=0}^\infty \frac {f^{(n)}(a)} {n!}(x-a)^n$$

where $$a = 0$$ is the Maclaurin series.

Wiki then states "The Maclaurin series for $$\frac 1 {1-x}$$ is the geometric series $$1 + x + x^2 + ...$$

My first question is, what is $$f(a)$$ for the Taylor Series that results in this MacLaurin series?

Wiki then goes into the Taylor series for $$\frac 1 x$$ at $$a = 1$$ is $$1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + ...$$

Does anyone know how this is derived?

If we take the function $$f(x) = 1/(1-x)$$ for $$|x|<1$$, then $$f^{(n)}(x) = n! /(1-x)^n$$ for all $$n$$ [by induction], so $$f^{(n)}(0) = n!$$ for all $$n$$, so the Maclaurin series is $$\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}(x-0)^n \sum_{n=0}^\infty \frac{n!}{n!}(x-0)^n= \sum_{n=0}^\infty x^n.$$

If we take $$f(x) = \frac{1}{x}$$ for $$x>0$$, then $$f^{(n)}(x) = \frac{(-1)^n n!}{x^{n+1}}$$ and $$f^{(n)}(1) = (-1)^n$$. The Taylor series at $$a=1$$ is $$\sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!}(x-1)^n = \sum_{n=0}^\infty \frac{(-1)^n n!}{n!}(x-1)^n = \sum_{n=0}^\infty (-1)^n(x-1)^n .$$

To start with, let us consider the geometric sequence whose first term equals one with ratio $$|x| < 1$$.

The we can express the sum of its first $$n+1$$ terms by \begin{align*} s_{n}(x) = 1 + x + x^{2} + \ldots + x^{n} & \Rightarrow xs_{n}(x) = x + x^{2} + x^{3} + \ldots + x^{n+1}\\\\ & \Rightarrow (1-x)s_{n}(x) = 1 - x^{n+1}\\\\ & \Rightarrow s_{n}(x) = \frac{1 - x^{n+1}}{1-x} \end{align*}

Since $$x^{n}\to 0$$ for $$|x| < 1$$ as $$n$$ approaches infinity, one concludes that \begin{align*} \lim_{n\to\infty}s_{n}(x) = \lim_{n\to\infty}\frac{1-x^{n+1}}{1-x} = \frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + \ldots \end{align*}

Consequently, the desired function is given by $$f(x) = \dfrac{1}{1-x}$$.

Based on such results, it is possible to give an answer to your second question as well.

More precisely, if $$|x - 1| < 1$$, we can conclude that

\begin{align*} \frac{1}{x} = \frac{1}{1 - (1-x)} = 1 + (1 - x) + (1-x)^{2} + (1-x)^{3} + \ldots \end{align*}

Hopefully this helps!

• Using $f(x) = \frac 1 {1-x}$ and taking $f'(x) = \frac 1 {(x-2)^2}$ and substituting into the Taylor series, I do not arrive at the second term $x$.
– Nick
Dec 16 '20 at 2:36
• The derivative of $f(x)$ is given by $$f'(x) = \frac{1}{(1-x)^{2}}$$ More generally, as pointed out by @GEdgar, one has that $$f^{(n)}(x) = \frac{n!}{(1-x)^{n}}$$ Dec 16 '20 at 2:37