# New Series from Old 3

Rational Functions and a “mistake”

A geometric series is one in which each term is found by multiplying the preceding term by the same number or expression. This number is called the common ratio, r. Geometric series converge if, and only if, $\left| r \right|<1$. If a geometric series converges, then the sum of the (infinite number of) terms is $\displaystyle \frac{{{a}_{1}}}{1-r}$ where a1 is the first term.

We can use this to write series for rational expressions.

Example 1. The series for $\displaystyle \frac{1}{1+{{x}^{2}}}$ that we assumed in the last post can be rewritten as $\displaystyle \frac{1}{1-\left( -{{x}^{2}} \right)}$.   This has the same form as the sum of the geometric series so we can write it as a geometric series with a1 = 1 and r = –x2 . The result is $\displaystyle \frac{1}{1+{{x}^{2}}}=1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+\cdots +{{\left( -1 \right)}^{2n-1}}{{x}^{2n-2}}+\cdots ,\quad -1

Example 2: $\displaystyle \frac{6x}{2+x}=\frac{2x}{1-\left( -\tfrac{x}{2} \right)}$. Letting ${{a}_{1}}=3x$ and $r=-\tfrac{x}{2}$ we have $\displaystyle \frac{6x}{2+x}=\frac{3x}{1-\left( -\tfrac{x}{2} \right)}=3x+3x\left( -\tfrac{x}{2} \right)+3x{{\left( -\tfrac{x}{2} \right)}^{2}}+3x{{\left( -\tfrac{x}{2} \right)}^{3}}+\cdots$ $\displaystyle \frac{6x}{2+x}=3x-\tfrac{3}{2}{{x}^{2}}+\tfrac{3}{4}{{x}^{3}}-\tfrac{3}{8}{{x}^{4}}+\tfrac{3}{16}{{x}^{5}}+\cdots +\tfrac{{{\left( -1 \right)}^{2n-1}}}{{{2}^{n-1}}}3{{x}^{n}}+\cdots$

The interval of convergence is $\left| -\tfrac{x}{2} \right|<1$  or $-2.

Many power series for rational functions can be obtained in this way.

An instructive “mistake.”

I have heard of students making an interesting mistake with this kind of problem. Instead of dividing by 2, they divide by x and arrive at $\displaystyle \frac{6x}{2+x}=\frac{6}{1-\left( -\tfrac{2}{x} \right)}$ and then write the series as $\displaystyle \frac{6x}{2+x}=\frac{6}{1-\left( -\tfrac{2}{x} \right)}=6+6\left( -\tfrac{2}{x} \right)+6{{\left( -\tfrac{2}{x} \right)}^{2}}+6{{\left( -\tfrac{2}{x} \right)}^{3}}+\cdots$ $\displaystyle \frac{6x}{2+x}=6-\frac{12}{x}+\frac{24}{{{x}^{2}}}-\frac{48}{{{x}^{3}}}+\frac{96}{{{x}^{4}}}-\cdots$

With an interval of convergence of $\left| -\tfrac{2}{x} \right|<1$ or $x<-2\text{ or }x>2$. Now this is not a Taylor series since the powers are in the denominators, but it is nevertheless interesting. Let’s look at the graphs. The rational expression is the black graph and partly hidden by the other graphs where they converge. The blue graph is the 5th degree Taylor Polynomial and its interval of convergence is the white strip in the center of the graph. The red graph is the “mistake” (also 5th degree) with the two blue regions as its interval of convergence.  Both series fit the rational function but only in their own interval of convergence.

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