Identify the Function Represented by the Following Power Series

Answer

Each series repeats its coefficient pattern with a fixed period, so each one can be split into a geometric series per residue class of the exponent and then summed with

$\sum_{k=0}^{\infty}r^{k}=\frac{1}{1-r}\qquad\text{for }|r|<1.$

79. $x+x^{2}-x^{3}+x^{4}+x^{5}-x^{6}+\cdots=\dfrac{x+x^{2}-x^{3}}{1-x^{3}}$ for $|x|<1$

The signs repeat every three terms. Grouping by exponent modulo $3$:

$x+x^{2}-x^{3}+x^{4}+x^{5}-x^{6}+\cdots=\left(x+x^{2}-x^{3}\right)+x^{3}\left(x+x^{2}-x^{3}\right)+x^{6}\left(x+x^{2}-x^{3}\right)+\cdots$

$=\left(x+x^{2}-x^{3}\right)\sum_{k=0}^{\infty}\left(x^{3}\right)^{k}=\frac{x+x^{2}-x^{3}}{1-x^{3}}.$

The geometric series has ratio $x^{3}$, so the identification holds for $|x^{3}|<1$, that is $|x|<1$.

80. $x+x^{2}-x^{3}-x^{4}+x^{5}+x^{6}-x^{7}-x^{8}+\cdots=\dfrac{x+x^{2}-x^{3}-x^{4}}{1-x^{4}}$ for $|x|<1$

Here the sign block $+,\ +,\ -,\ -$ repeats every four terms:

$=\left(x+x^{2}-x^{3}-x^{4}\right)\sum_{k=0}^{\infty}\left(x^{4}\right)^{k}=\frac{x+x^{2}-x^{3}-x^{4}}{1-x^{4}}.$

The numerator factors as $x(1+x)(1-x^{2})$ and the denominator as $(1-x)(1+x)(1+x^{2})$, so cancelling the common factor $(1-x)(1+x)$ gives the equivalent closed form

$\frac{x+x^{2}-x^{3}-x^{4}}{1-x^{4}}=\frac{x(1+x)}{1+x^{2}},\qquad |x|<1.$

Both forms describe the same function on the interval of convergence; the second is shorter, the first shows the grouping directly.

81. $x-x^{2}-x^{3}+x^{4}-x^{5}-x^{6}+x^{7}-\cdots=\dfrac{x-x^{2}-x^{3}}{1-x^{3}}$ for $|x|<1$

The sign block is now $+,\ -,\ -$, and the same three-term grouping applies:

$=\left(x-x^{2}-x^{3}\right)\sum_{k=0}^{\infty}\left(x^{3}\right)^{k}=\frac{x-x^{2}-x^{3}}{1-x^{3}},\qquad |x|<1.$

Reading a repeated pattern

The three answers come from one method:

1. Find the period $m$ of the sign pattern and write down the terms of one full period.
2. Factor that block out of every later period: the remaining factor is $\left(x^{m}\right)^{k}$.
3. Sum the geometric series and state the interval where its ratio satisfies $|x^{m}|<1$.

A quick numerical check on 79 at $x=\frac{1}{2}$: the closed form gives $\frac{0.5+0.25-0.125}{1-0.125}=\frac{0.625}{0.875}\approx 0.714$, and the partial sums $0.5,\ 0.75,\ 0.625,\ 0.6875,\ 0.71875,\ \dots$ move toward the same value. Checking one point like this catches a wrong sign block immediately.

Outside the interval of convergence the identification fails: the rational function is still defined at, say, $x=2$, but the series no longer converges there, so the two are not equal.