Although the section number is from our official course textbook, Stewart Calculus: Early Transcendentals, these examples are from the (vastly superior) textbook, Thomas Calculus, Twelfth Edition. Its author, George Thomas, was an MIT professor and a master of explaining calculus concepts.
How to test a power series for convergence by using the _Ratio Test_ to determine for which values of $x$ it converges/diverges.
For what values of $x$ do the following power series converge?
a) $\quad \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - ... + \frac{x^n}{n}$
First, examine a larger list of terms:
$$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n} = x - \frac{x^2}{n} + \frac{x^3}{n} - \frac{x^4}{4} + \frac{x^5}{n} - \frac{x^6}{n} + \frac{x^7}{7} - ...$$The Ratio Test says that for series with positive terms, you take the limit of the ratio of two consecutive terms. If the result, $p$ is less than one, the series converges:
$$ \require{cancel} \color{blue}p\color{black} = \lim_\limits{n \to \infty} \left|\frac{a_{n+1}}{a_n} \right| \\ \text{If } \color{blue}p\color{black} < 1 \text{, the series converges.} \\ \text{If } \color{blue}p\color{black} > 1 \text{ or } \color{blue}p\color{black} \to \infty \text{ as } n \to \infty \text{, the series diverges.} \\ \text{If } \color{blue}p\color{black}=1 \text{, the result is inconclusive, meaning } \\ \text{you can't tell if the series is convergent or divergent.} $$For the series in a),, the ratio test, is:
$$ \lim_\limits{n \to \infty} \left| \frac{ \frac{x^{n+1}}{n+1} }{ \frac{x^n}{n} } \right| = \color{blue}p $$$$ = \lim_\limits{n \to \infty} \left| \frac{x^{n+1}}{n+1} \cdot \frac{n}{x^n} \right| = \lim_\limits{n \to \infty} \left| \frac{x^{\cancel{n+1}}}{n+1} \cdot \frac{n}{\cancel{x^n}} \right|$$$$ = \lim_\limits{n \to \infty} \frac{n}{n+1} \cdot |x| $$$$ = \left( \frac{n}{n} \right) \cdot |x| \quad \text{ (<= n / (n+1) behaves like 1 as } n \to \infty \text{)} $$$$= |x|$$So, the series converges for $|x| < 1$, which is the same as saying, $-1 < x < 1$.
It diverges if $|x| > 1$, because in that scenario the nth term does not converge to zero.
Therefore, the radius of convergence $R$ is equal to one: $R = 1$.
b) $\quad \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{2n-1}}{2n-1} = \left[ (-1)^{1-1} \cdot \frac{x^{2(1)-1}}{2(1)-1} \right] + \left[ (-1)^{2-1} \cdot \frac{x^{2(2)-1}}{2(2)-1} \right] + \left[ (-1)^{3-1} \cdot \frac{x^{2(3)-1}}{2(3)-1} \right]$
$$= \sum_{n=1}^{\infty} \left( (-1)^{n-1} \cdot \frac{x^{2n-1}}{2n-1} \right) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^8}{8} - ... $$The Ratio Test for this power series looks like this:
$$ \color{blue}p\color{black} = \lim_\limits{n \to \infty} \left| \frac{ \frac{x^{2(n+1)-1}}{2(n+1)-1} }{ \frac{x^{2n-1}}{2n-1} } \right| = \lim_\limits{n \to \infty} \left| \frac{x^{2(n+1)-1}}{2(n+1)-1} \cdot \frac{2n-1}{x^{2n-1}} \right| $$
In [ ]: