A power series is an infinite series of the form $\sum_{n=0}^\infty c_n (x-a)^n$, where $c_n$ are coefficients and $a$ is the center. Power series can represent functions like $e^x$, $\sin x$, and $\frac{1}{1-x}$ within their interval of convergence. The set of $x$ values for which the series converges is an interval centered at $a$.
🎯 In this section you will learn
- The definition of a power series and its center
- How to find the radius and interval of convergence
- Using the Ratio Test and Root Test to find convergence
- Checking endpoints for convergence
📌 Definition
$$ \sum_{n=0}^\infty c_n (x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots $$
Power series centered at $x=a$
💡 Key Facts
- A power series always converges at $x=a$ (all terms except $c_0$ vanish).
- The set of convergence is an interval centered at $a$: $(a-R, a+R)$, possibly including endpoints.
- $R$ is called the radius of convergence (can be $0$ or $\infty$).
📌 Finding the Radius of Convergence
Use the Ratio Test or Root Test:
$$ R = \frac{1}{\lim_{n \to \infty} |c_{n+1}/c_n|^{1/n}} \quad \text{or} \quad \frac{1}{\lim_{n \to \infty} |c_n|^{1/n}} $$
Find the radius and interval of convergence for $\sum_{n=0}^\infty \frac{x^n}{n!}$.
①
Apply Ratio Test
$\lim_{n \to \infty} \left| \frac{x^{n+1}/(n+1)!}{x^n/n!} \right| = \lim_{n \to \infty} \frac{|x|}{n+1} = 0$
②
Conclusion
The limit is $0$ for all $x$, so $R = \infty$. The series converges for all real $x$.
Find the radius and interval of convergence for $\sum_{n=1}^\infty \frac{(x-2)^n}{n}$.
①
Apply Ratio Test
$\lim_{n \to \infty} \left| \frac{(x-2)^{n+1}/(n+1)}{(x-2)^n/n} \right| = \lim_{n \to \infty} |x-2| \cdot \frac{n}{n+1} = |x-2|$
②
Convergence condition
$|x-2| < 1$ → $1 < x < 3$. Radius $R = 1$.
③
Check endpoints
At $x=3$: $\sum 1/n$ (harmonic) diverges. At $x=1$: $\sum (-1)^n/n$ (alternating harmonic) converges.
④
Interval of convergence
$[1, 3)$
Find the radius of convergence for $\sum_{n=0}^\infty n! x^n$.
- Ratio Test: $\lim \frac{(n+1)! |x|^{n+1}}{n! |x|^n} = \lim (n+1)|x| = \infty$ for $x \neq 0$.
- Converges only at $x=0$. Radius $R = 0$.
Find the interval of convergence for $\sum_{n=0}^\infty x^n$.
- Geometric series: converges when $|x| < 1$. At $x=1$: diverges; at $x=-1$: diverges (alternates but terms don't go to 0).
- Interval: $(-1, 1)$.
⚠️ Common Mistakes
- Forgetting to check endpoints: The ratio test only gives the open interval; endpoints must be checked separately.
- Misidentifying the center: In $\sum c_n (x-a)^n$, $a$ is the center. For $\sum c_n x^n$, the center is $0$.
- Confusing radius with interval: Radius $R$ tells how far from $a$ the series converges; interval includes endpoints if they converge.
🔍 Key Takeaways
- Power series: $\sum c_n (x-a)^n$.
- Radius of convergence $R$: converges for $|x-a| < R$, diverges for $|x-a| > R$.
- Use Ratio Test or Root Test to find $R$.
- Always check endpoints separately to determine the interval of convergence.
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