Section 8: Power Series
A power series is a function of the form $f(x) = \sum_{n=0}^\infty c_nx^n$ (or $\sum_{n=0}^\infty c_n(x-a)^n$, which is said to be centered at $a$), which converges for some values of $x$. It will converge on an interval of convergence. This interval might just be a point (for example, $[2, 2] = \{2\}$); it might be an open or closed or half-open interval (for example, $(0, 2), [-1, 1], [2, 5)$). Or it might be the entire real line $\mathbb{R} = (-\infty, \infty)$. The radius of convergence $R$ is half the length of the interval.
The ratio test is typically what's used to find the radius of convergence for a power series. Once you have the radius, you must check the endpoints individually. The next example illustrates this.
Example: Let's use the ratio test to find the interval of convergence for $f(x) = \sum_{n=1}^\infty \frac{(2x-1)^n}{n 5^n}$. The ratio is $|2x-1|/5$. For this to be convergent by the ratio test, it must be $< 1$. This is equivalent to $|2x - 1| < 5$, or $|x - 1/2| < 5/2$. So $-2 < x < 3$. (The radius of convergence is $5/2$.) Remember that the ratio test doesn't say what happens on the boundary, so we have to test the cases $x = -2$ and $x = 3$ separately.
If $x = -2$, the series is $\sum_{n=1}^\infty \dfrac{(-5)^n}{n 5^n} = \sum_{n=1}^\infty \dfrac{(-1)^n}{n}$, which is the convergent alternating harmonic series. If $x = 3$, the series is $\sum_{n=1}^\infty \dfrac{5^n}{n 5^n} = \sum_{n=1}^\infty \dfrac{1}{n}$, which is the divergent harmonic series. So the interval of convergence is $[-2, 3)$.
Mini-Quiz: Compute the interval of convergence for the power series $\sum_{n=1}^\infty \dfrac{(x-1)^n(n+3)}{7^n}$.
Now let's figure out how to write a general function as a power series. We will see in the next section how to use Taylor series formulas to do this, but for now, our approach will be to manipulate $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$ (which converges for $|x| < 1$) into the series we want. In particular, the most powerful thing we can do is differentiate or integrate series we already know.
Example: Let's differentiate both sides of the equation $1/(1-x) = \sum_{n=0}^\infty x^n$. We find $1/(1-x)^2 = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1} = \sum_{n=0}^\infty (n+1)x^n$.
When we differentiate or integrate power series, we keep the same ratio of convergence. But convergence should always be checked at the endpoints.
Mini-Quiz: Integrate $1/(1-x) = \sum_{n=0}^\infty x^n$. What do you get?
Example: One important example is to find a power series for $\arctan{x}$. The idea is to look at the derivative first. The derivative of $\arctan{x}$ is $\frac{1}{1 + x^2} = \frac{1}{1 - (-x^2)}$. We can use the fact that $\frac{1}{1 - x} = \sum_{n=0}^\infty x^n, |x| < 1$, to write $\frac{1}{1 - (-x^2)} = \sum_{n=0}^\infty (-x^2)^n = \sum_{n=0}^\infty (-1)^nx^{2n}$. The interval of convergence is $|-x^2| < 1$, which is equivalent to $|x| < 1$.
We then pull back and integrate to get a series for $\arctan{x}$. $\arctan{x} = \int \frac{1}{1+x^2}\;dx = \int \sum_{n=0}^\infty(-1)^nx^{2n}\;dx = \sum_{n=0}^\infty \dfrac{(-1)^nx^{2n+1}}{2n+1} + C$ for some constant $C$. Plug in $0$ to both sides to find $C$: $\arctan{0} = 0 = C$. So $$\arctan{x} = \sum_{n=0}^\infty \dfrac{(-1)^nx^{2n+1}}{2n+1}.$$ This is valid for $|x| < 1$, but not for $|x| > 1$ because it is derived from the formula for a geometric series.
Mini-Quiz: What happens at $|x| = 1$? Does the power series for $\arctan{x}$ converge at $x = \pm 1$?