Power Series: Radius & Interval of Convergence
Section 6.1: how to work the HW 6.1 problems
A power series centered at \(a\) is a series with a variable in it:
\[ \sum_{n=0}^\infty c_n (x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots \]
For each number you plug in for \(x\), you get an ordinary series that either converges or diverges. The whole job this week is answering: for which \(x\) does it converge?
1. Two facts that are always true (#1)
A power series always converges at its center. At \(x = a\), every term except the first is \(0\):
\[ \sum c_n (a-a)^n = c_0 + 0 + 0 + \cdots = c_0 \]
For \(\sum a_n x^n\) the center is \(0\), so it converges at \(x = 0\) no matter what the \(a_n\) are. (If the sum starts at \(n = 1\), every term is \(0\) and the sum is \(0\).)
If it converges at some \(x\), then its terms go to \(0\) there. This is just the Divergence Test: if \(\sum a_n x^n\) converges, then \(a_n x^n \to 0\).
2. Radius of convergence from the ratio test (#2–4)
For any power series \(\sum c_n (x-a)^n\), exactly one of these is true:
- it converges only at \(x = a\) (radius \(R = 0\));
- it converges for all \(x\) (radius \(R = \infty\));
- there is a number \(R > 0\) so that it converges for \(|x - a| < R\) and diverges for \(|x - a| > R\).
\(R\) is the radius of convergence. At the two endpoints \(x = a \pm R\), anything can happen.
How to find \(R\): run the ratio test on the absolute values, keeping \(x\) as a fixed number:
\[ \lim_{n\to\infty}\left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right| = |x-a|\cdot\underbrace{\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|}_{L} \]
The series converges when \(|x - a|\cdot L < 1\), that is, when \(|x - a| < \frac1L\). So
\[ R = \frac{1}{L}, \qquad L = \lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right| \qquad (L = 0 \Rightarrow R = \infty,\quad L = \infty \Rightarrow R = 0) \]
If the coefficient ratio goes to \(L = 4\), then \(R = \frac14\), not \(4\). Solving \(|x|\cdot L < 1\) for \(|x|\) every time avoids this mistake.
The “scaling” shortcut
Suppose all you know is \(\left|\frac{a_{n+1}}{a_n}\right|\to 1\), so \(\sum a_n x^n\) has \(R = 1\). What if each term is also scaled?
Example. \(\displaystyle\sum_{n=0}^\infty \frac{a_n x^n}{5^n}\).
\[ \left|\frac{a_{n+1}x^{n+1}/5^{n+1}}{a_n x^n/5^n}\right| = \left|\frac{a_{n+1}}{a_n}\right|\cdot\frac{|x|}{5} \longrightarrow 1\cdot\frac{|x|}{5} \]
This converges when \(\frac{|x|}{5} < 1\), so \(R = 5\).
- A sign factor \((-1)^n\) changes nothing, because the ratio test uses absolute values.
- Example. \(\displaystyle\sum a_n \frac{2^n}{e^n}x^n\): the ratio goes to \(\frac{2}{e}|x|\), so \(R = \frac{e}{2} \approx 1.359\).
If \(\left|\frac{a_{n+1}}{a_n}\right|\to1\), then \(\displaystyle\sum a_n\, b^n x^n\) has \(R = \frac{1}{|b|}\). Dividing by \(c^n\) means \(b = \frac1c\), so \(R = c\).
3. The full method: radius and interval (#5–8)
- Ratio test on \(\left|\frac{a_{n+1}}{a_n}\right|\), keeping \(x\) in it.
- Solve “limit \(< 1\)” for \(x\). This gives \(|x - a| < R\).
- Check both endpoints \(x = a - R\) and \(x = a + R\) by plugging them in. Each gives a plain number series; use any earlier test.
- Write the interval with brackets for endpoints that converge and parentheses for endpoints that diverge.
The ratio test gives exactly \(1\) at the endpoints, so it can never decide them. You must plug them in.
Example A (one endpoint in). \(\displaystyle\sum_{n=1}^\infty \frac{(3x)^n}{\sqrt n}\)
Step 1: \[ \left|\frac{(3x)^{n+1}/\sqrt{n+1}}{(3x)^n/\sqrt n}\right| = 3|x|\sqrt{\frac{n}{n+1}} \longrightarrow 3|x| \]
Step 2: \(3|x| < 1 \iff |x| < \frac13\), so \(R = \frac13\).
Step 3:
- \(x = \frac13\): \(\displaystyle\sum\frac{1^n}{\sqrt n} = \sum\frac{1}{\sqrt n}\), a \(p\)-series with \(p = \frac12\). Diverges.
- \(x = -\frac13\): \(\displaystyle\sum\frac{(-1)^n}{\sqrt n}\). AST: \(\frac{1}{\sqrt n}\) decreases to \(0\). Converges (conditionally).
Step 4: \(\boxed{\left[-\tfrac13,\ \tfrac13\right)}\)
Example B (neither endpoint). \(\displaystyle\sum_{n=1}^\infty \frac{n\,x^n}{4^n}\)
\[ \left|\frac{(n+1)x^{n+1}/4^{n+1}}{n\,x^n/4^n}\right| = \frac{n+1}{n}\cdot\frac{|x|}{4}\longrightarrow\frac{|x|}{4}, \qquad R = 4 \]
- \(x = 4\): \(\sum n\). Terms \(\to\infty\). Diverges.
- \(x = -4\): \(\sum (-1)^n n\). Terms don’t \(\to 0\). Diverges.
Interval: \(\boxed{(-4,\ 4)}\). Powers of \(n\) in the coefficient, like \(n\) or \(n^2\), never change \(R\); they only affect the endpoints.
Example C (converges everywhere). \(\displaystyle\sum_{k=0}^\infty \frac{5^k x^k}{k!}\)
\[ \left|\frac{5^{k+1}x^{k+1}/(k+1)!}{5^k x^k/k!}\right| = \frac{5|x|}{k+1}\longrightarrow 0 < 1 \quad\text{for every } x \]
So \(R = \infty\) and the interval is \(\boxed{(-\infty,\ \infty)}\). There are no endpoints to check. A factorial in the denominator beats any exponential.
Example D (not centered at 0). \(\displaystyle\sum_{n=1}^\infty \frac{(x-2)^n}{n\,3^n}\)
\[ \left|\frac{(x-2)^{n+1}/((n+1)3^{n+1})}{(x-2)^n/(n\,3^n)}\right| = \frac{n}{n+1}\cdot\frac{|x-2|}{3}\longrightarrow\frac{|x-2|}{3} \]
\(|x - 2| < 3\), so \(R = 3\) and the endpoints are \(2 \pm 3\), that is, \(-1\) and \(5\).
- \(x = 5\): \(\sum\frac{3^n}{n\,3^n} = \sum\frac1n\). Diverges.
- \(x = -1\): \(\sum\frac{(-3)^n}{n\,3^n} = \sum\frac{(-1)^n}{n}\). Converges (AST).
Interval: \(\boxed{[-1,\ 5)}\).
Type intervals like [-1,5), (-4,4), or (-inf,inf). Decimals are accepted for messy endpoints, but exact values like 1/3 are safer.
4. Factorial coefficients (#9, 11, 12)
Same method; the work is all in simplifying the ratio. Two pieces of bookkeeping:
- \((kn)!\) gains \(k\) factors when \(n \to n+1\). For example, \((2n+2)! = (2n+2)(2n+1)\,(2n)!\) and \((3n+3)! = (3n+3)(3n+2)(3n+1)\,(3n)!\).
- \(n^n\) is not a factorial. \(\frac{(n+1)^{n+1}}{n^n} = (n+1)\left(\frac{n+1}{n}\right)^n\), and \(\left(1 + \frac1n\right)^n\to e\).
Example. \(\displaystyle\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2}x^n\)
\[ \left|\frac{c_{n+1}}{c_n}\right| = \frac{(2n+2)!}{\big((n+1)!\big)^2}\cdot\frac{(n!)^2}{(2n)!} = \frac{(2n+2)(2n+1)}{(n+1)^2} = \frac{2(2n+1)}{n+1} \longrightarrow 4 \]
So \(L = 4\) and \(R = \frac14\).
Example. \(\displaystyle\sum_{n=1}^\infty \frac{n^n}{n!}x^n\)
\[ \left|\frac{c_{n+1}}{c_n}\right| = \frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n} = \frac{(n+1)^{n+1}}{(n+1)\,n^n} = \left(\frac{n+1}{n}\right)^n = \left(1+\frac1n\right)^n \longrightarrow e \]
So \(L = e\) and \(R = \frac1e\). (Flip the fraction, \(\frac{n!}{n^n}\), and you get \(L = \frac1e\) and \(R = e\).)
Extra constants just multiply \(L\). For example, \(\sum \frac{9^n(n!)^2}{(2n)!}x^n\) has \(L = 9\cdot\frac14\), so \(R = \frac49\).
5. When the ratio test breaks (#10)
Coefficients like \(\sin^2(k)\) or \(\cos(k)\) bounce around, and some can even be \(0\). Then \(\frac{c_{k+1}}{c_k}\) is undefined or has no limit. Use comparison instead.
Example. \(\displaystyle\sum_{k=1}^\infty \big(2 + \sin k\big)\,x^k\). The coefficients are trapped: \(1 \le 2 + \sin k \le 3\).
- \(|x| < 1\): \(\big|(2+\sin k)x^k\big| \le 3|x|^k\), and \(\sum 3|x|^k\) is a convergent geometric series. Converges (absolutely).
- \(|x| \ge 1\): \(\big|(2+\sin k)x^k\big| \ge 1\cdot|x|^k \ge 1\). The terms don’t go to \(0\). Diverges.
So \(R = 1\).
If the coefficients stay between two positive constants, the power series behaves like \(\sum x^k\), so \(R = 1\). For \(\sin^2 k\) the lower bound is \(0\), not positive, so you need one more fact, proved in the HW 5.5 solutions: \(\sin^2 k \not\to 0\). That shows divergence at \(|x| \ge 1\).
6. New series from \(\frac{1}{1-x}\) (#13–17)
Start from the one series we know a closed form for:
\[ \frac{1}{1-u} = \sum_{n=0}^\infty u^n, \qquad |u| < 1 \]
Get the function into the form \((\text{stuff})\cdot\dfrac{1}{1 - u}\), replace \(u\) by whatever expression you have, then find the interval from \(|u| < 1\).
Substitute a power. \(\dfrac{1}{1+x^2} = \dfrac{1}{1-(-x^2)}\), so \(u = -x^2\):
\[ \frac{1}{1+x^2} = \sum_{n=0}^\infty (-x^2)^n = \sum_{n=0}^\infty (-1)^n x^{2n}, \qquad |x^2| < 1 \iff x \in (-1, 1) \]
Multiply by a power out front. \(\dfrac{x^2}{1-x^5} = x^2\cdot\dfrac{1}{1-x^5}\):
\[ x^2\sum_{n=0}^\infty x^{5n} = \sum_{n=0}^\infty x^{5n+2}, \qquad x\in(-1,1) \]
Substitute a multiple. \(\dfrac{1}{1-4x}\) with \(u = 4x\):
\[ \frac{1}{1-4x} = \sum_{n=0}^\infty (4x)^n = \sum_{n=0}^\infty 4^n x^n, \qquad |4x| < 1 \iff x\in\left(-\tfrac14,\ \tfrac14\right) \]
Both at once. \(\dfrac{3x}{1-2x^2} = 3x\cdot\dfrac{1}{1-2x^2}\) with \(u = 2x^2\):
\[ 3x\sum_{n=0}^\infty (2x^2)^n = \sum_{n=0}^\infty 3\cdot 2^n x^{2n+1}, \qquad 2x^2 < 1 \iff |x| < \frac{1}{\sqrt2} \]
So the interval is \(\left(-\frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right) \approx (-0.7071, 0.7071)\).
After substituting \(u = 4x\) or \(u = 2x^2\), the interval shrinks or stretches. And the endpoints are never included for these geometric series: at \(|u| = 1\) the terms don’t go to \(0\).
Recentering: series in powers of \((x - a)\)
To expand about \(a \ne 0\), rewrite everything in terms of \((x - a)\).
Easy case. \(\dfrac{1}{4-x}\) centered at \(a = 3\). Write \(4 - x = 1 - (x - 3)\):
\[ \frac{1}{4-x} = \frac{1}{1-(x-3)} = \sum_{n=0}^\infty (x-3)^n, \qquad |x - 3| < 1 \iff x\in(2,\ 4) \]
General case. \(\dfrac{1}{5-x}\) centered at \(a = 2\). Now \(5 - x = 3 - (x-2)\), and there’s a \(3\) where we need a \(1\), so factor it out:
\[ \frac{1}{5-x} = \frac{1}{3 - (x-2)} = \frac13\cdot\frac{1}{1 - \frac{x-2}{3}} = \frac13\sum_{n=0}^\infty\left(\frac{x-2}{3}\right)^n = \sum_{n=0}^\infty\frac{(x-2)^n}{3^{n+1}} \]
with \(\left|\frac{x-2}{3}\right| < 1 \iff |x-2| < 3 \iff x\in(-1,\ 5)\).
- Write the denominator as \((\text{number}) - (x - a)\).
- Factor out the number so the denominator reads \(1 - \frac{x-a}{\text{number}}\).
- Substitute \(u = \frac{x-a}{\text{number}}\) into \(\sum u^n\).
- The interval is \(|x - a| < |\text{number}|\), centered at \(a\).
Summary checklist
| Problem type | HW 6.1 | Key move |
|---|---|---|
| True/false basics | #1 | always converges at the center; terms \(\to0\) if convergent |
| “Suppose \(\lvert a_{n+1}/a_n\rvert\to1\)” | #2–4 | extra factor \(b^n\) gives \(R = 1/\lvert b\rvert\) |
| Radius + interval | #5–8 | ratio test, solve for \(\lvert x-a\rvert\), check endpoints |
| Factorials / \(n^n\) | #9, 11, 12 | \((kn)!\) gains \(k\) factors; \(\left(1+\frac1n\right)^n\to e\); \(R = 1/L\) |
| Bouncing coefficients | #10 | bound the coefficients and compare with a geometric series |
| From \(\frac{1}{1-x}\) | #13–17 | substitute \(u\), multiply by prefactor, interval from \(\lvert u\rvert<1\); recenter by factoring |