Manipulating Power Series
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 6.2: “Properties of Power Series” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/6-2-properties-of-power-series |
| Textbook used in class | Stewart, Calculus, Section 11.9: “Representations of Functions as Power Series” (Examples 4, 5, 6) |
Opening Scenario
If a function has a power series representation, you can differentiate or integrate that series term by term, just as you would a polynomial. The result is the power series for the derivative or the antiderivative. This lets you build series for $\ln(1+x)$, $\arctan x$, and other functions by integrating the geometric series, without needing to compute derivatives of the function directly.
Quick Reference
Theorem. If $f(x) = \displaystyle\sum_{n=0}^{\infty} c_n(x-a)^n$ has radius of convergence $R > 0$, then on the interval $(a-R, a+R)$:
Differentiation term by term: \[ f'(x) = \sum_{n=1}^{\infty} n\, c_n (x-a)^{n-1}. \]
Integration term by term: \[ \int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1}. \]
Both derived series have the same radius of convergence $R$. The behavior at the endpoints of the interval may differ.
Key Concepts
1. Differentiation to Find a New Series
Example 1. Find a power series for $\dfrac{1}{(1-x)^2}$ and its interval of convergence. (Stewart 11.9, Example 4.)
Start from $\dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ and differentiate both sides: \[ \frac{d}{dx}\left[\frac{1}{1-x}\right] = \frac{1}{(1-x)^2}, \qquad \frac{d}{dx}\left[\sum_{n=0}^{\infty} x^n\right] = \sum_{n=1}^{\infty} nx^{n-1}. \] So: \[ \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1} = \sum_{n=0}^{\infty}(n+1)x^n. \]
Boxed answer: $\dfrac{1}{(1-x)^2} = \displaystyle\sum_{n=1}^{\infty} nx^{n-1}$, for $|x| < 1$.
2. Integration to Find a New Series
Example 2. Find a power series for $\ln(1+x)$ centered at $0$. (Stewart 11.9, Example 5.)
Start from $\dfrac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n$ (valid for $|x| < 1$). Integrate both sides from $0$ to $x$: \[ \int_0^x \frac{dt}{1+t} = \sum_{n=0}^{\infty} (-1)^n \int_0^x t^n\,dt. \] \[ \ln(1+x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]
The constant of integration is $0$ because $\ln(1+0) = 0$ and the series is $0$ at $x = 0$.
At $x = 1$: the series becomes the alternating harmonic series $\sum (-1)^n/(n+1)$, which converges. This gives the beautiful result $\ln 2 = 1 - 1/2 + 1/3 - 1/4 + \cdots$.
Boxed answer: $\ln(1+x) = \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1} x^n}{n}$, for $-1 < x \leq 1$.
Example 3. Find a power series for $\arctan x$ centered at $0$. (Stewart 11.9, Example 6.)
$\arctan'(x) = 1/(1+x^2) = \sum_{n=0}^\infty (-1)^n x^{2n}$. Integrate from $0$ to $x$: \[ \arctan x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \] At $x = 1$: $\arctan 1 = \pi/4 = 1 - 1/3 + 1/5 - 1/7 + \cdots$ (Leibniz formula for $\pi$).
Boxed answer: $\arctan x = \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{2n+1}$, for $|x| \leq 1$.
differentiating or integrating changes the radius of convergence. The radius $R$ is unchanged by term-by-term differentiation or integration. What may change is the behavior at the endpoints: a series that converges at an endpoint might no longer converge there after differentiation (since differentiation makes endpoint convergence harder), or might start converging there after integration. Check the endpoint behavior separately for each derived series.
Common Errors Summary
| Error | Correction |
|---|---|
| Forgetting the constant of integration when integrating | Determine $C$ by plugging in a known value (usually $x = 0$) |
| Shifting the index incorrectly after differentiating | After $\sum n c_n x^{n-1}$, re-index by letting $m = n-1$: $\sum (m+1)c_{m+1}x^m$ |
| Claiming the interval changes after differentiation or integration | The radius $R$ is the same; only endpoint inclusion may change |
Leveled Practice
Problem 1. Use differentiation of $1/(1-x)$ to find a series for $\dfrac{x}{(1-x)^2}$.
Show answer
$\dfrac{1}{(1-x)^2} = \sum_{n=1}^\infty n x^{n-1}$. Multiply both sides by $x$: $\dfrac{x}{(1-x)^2} = \sum_{n=1}^\infty n x^n$, for $|x| < 1$.
Problem 2. Evaluate $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1) \cdot 3^{2n+1}}$.
Show answer
Use $\arctan x = \sum_{n=0}^\infty (-1)^n x^{2n+1}/(2n+1)$ and set $x = 1/3$: \[ \arctan(1/3) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1) \cdot 3^{2n+1}}. \]
The series sums to $\arctan(1/3)$.
Mastery Checklist
Previous: Power Series Representations | Next: Taylor Series