Recheck the convergence interval
One series, then its derivatives
A power series with radius of convergence \(R\) converges at every \(x\) closer to its center than \(R\). At an endpoint, the center plus or minus \(R\), it may converge or diverge.
The teal bar on the first line is the interval of \(f(x)=\sum_{n=1}^{\infty}\dfrac{x^n}{n^22^n}\), with center 0 and radius 2. A filled ring means that the series converges at that end, and an open ring means that it diverges. The lower graph shows the partial sums at the orange point.
At \(x=2\), the terms are \(\frac1{n^2}\), and the partial sums approach a limit. At \(x=-2\), the terms have the same sizes, so both rings are filled.
Press Differentiate. The radius is still 2. At \(x=2\), the new terms are \(\frac1{2n}\), and the partial sums grow without a limit, so that ring is open. At \(x=-2\), the terms alternate and decrease to 0, so that ring stays filled.
Press it again. At both ends, the sizes of the terms approach \(\frac14\), not 0, so both rings are open. Press Integrate. At the ends, the terms of the antiderivative are no larger than \(\frac1{n^2}\), so its rings are filled.
Any power series
The center is now \(a\) and the radius is \(R\), so the teal bar runs from \(a-R\) to \(a+R\).
Press the buttons to add the series of the derivative and of an antiderivative. Each new bar has the same width, because the radius stays \(R\).
The terms at an end change, so its ring can change, and each end needs a new test.
Notation
Differentiating or integrating each term gives a series with the same center and radius.
\[\begin{aligned} &\sum c_n(x-a)^n \\ &\sum nc_n(x-a)^{n-1} \\ &\sum\dfrac{c_n(x-a)^{n+1}}{n+1} \end{aligned}\]Inside the interval, the new series converges, so the partial sums at the orange point approach a limit.
\[a-R\lt x\lt a+R\]Substitute each end for \(x\) and test the series of numbers that results.
\[x=a-R\quad\text{and}\quad x=a+R\]Write a bracket at an end where the series converges and a parenthesis where it diverges. The interval below includes only the left end, like that of \(f'\).
\[\cy{[a-R,\ a+R)}\]
When it fails
Copy the endpoints of \(f\) onto \(f'\). The ring of \(f'\) at 2 becomes filled, which is false.
At \(x=2\), the terms of \(f'\) are \(\frac1{2n}\), so its partial sums grow without a limit.
Example 1: the interval of a derivative
The series \(g(x)=\sum_{n=1}^{\infty}\dfrac{(x-3)^n}{n\cdot4^n}\) has center 3, radius 4 and interval of convergence \([-1,7)\). Find the interval of convergence of \(g'\).
Differentiate each term. The power rule brings down a factor \(n\), which cancels the \(n\) in the denominator.
\[g'(x)=\sum_{n=1}^{\infty}\dfrac{(x-3)^{n-1}}{4^n}\]The radius is still 4, so the new series converges for \(-1\lt x\lt 7\).
Test \(x=7\). Each term is \(\frac14\), so the terms do not approach 0, and the series diverges.
\[\dfrac{(7-3)^{n-1}}{4^n}=\dfrac{4^{n-1}}{4^n}=\dfrac14\]Test \(x=-1\). The terms \(\frac{(-4)^{n-1}}{4^n}\) have the size \(\frac14\) and do not approach 0, so the series diverges, and its partial sums jump between \(\frac14\) and 0.
The end \(x=-1\) converged for \(g\), but both ends diverge for \(g'\).
\[\cy{(-1,\ 7)}\]
Example 2: the interval of an integral
The series \(\sum_{n=0}^{\infty}(2x)^n\) has radius \(\frac12\). Integrate it term by term, with the constant 0, and find the interval of convergence of the new series.
The series is geometric with ratio \(2x\), so it converges for \(|2x|<1\). At \(x=\pm\frac12\), the terms do not approach 0, so both ends diverge, as the partial sums show.
Integrate each term. The radius is still \(\frac12\), and the new series sums to the antiderivative of \(\frac1{1-2x}\) that is 0 at \(x=0\).
\[\sum_{n=0}^{\infty}\dfrac{2^nx^{n+1}}{n+1}=-\dfrac12\ln(1-2x)\]A tempting wrong move: Both ends failed for the original series, so keep the open interval.
\[\left(-\tfrac12,\ \tfrac12\right)\]At \(x=-\frac12\), the partial sums approach the dashed line at \(-\frac12\ln2\), about \(-0.347\), so the series converges at that end.
At \(x=-\frac12\), the terms \(\frac{(-1)^{n+1}}{2(n+1)}\) alternate and decrease to 0, so the series converges. At \(x=\frac12\), the terms \(\frac1{2(n+1)}\) are half those of the harmonic series, so it diverges.
\[\cy{\left[-\tfrac12,\ \tfrac12\right)}\]
Practice
The series \(k(x)=\sum_{n=1}^{\infty}\dfrac{x^n}{n^3}\) has radius 1. Find the interval of convergence of \(k'\).
Test the series of \(k'\) at each end of its bar, \(-1\) and 1.
Solution
Differentiate each term. The factor \(n\) reduces \(n^3\) to \(n^2\).
\[k'(x)=\sum_{n=1}^{\infty}\dfrac{x^{n-1}}{n^2}\]The radius is still 1, so \(k'\) converges for \(-1\lt x\lt 1\).
At \(x=1\), the terms \(\frac1{n^2}\) form a p-series with \(p=2>1\), so it converges.
At \(x=-1\), the terms have the same sizes \(\frac1{n^2}\), so this series converges too.
Both ends converge, so both stay in the interval.
\[[-1,\ 1]\]
Questions
The number lines show the intervals of \(u(x)=\sum_{n=1}^{\infty}\dfrac{(x+1)^n}{n\cdot3^n}\), of the series of \(u'\) and of the series of an antiderivative of \(u\). A filled ring means that the series converges at that end.
Which end does the series of \(u'\) lose that the series of \(u\) has?
- \(x=-4\)
- \(x=2\)
- Neither end
Answer
\(x=-4\)
What is the interval of convergence of the series of the antiderivative?
- \([-4,\ 2]\)
- \((-4,\ 2)\)
- \([-4,\ 2)\)
Answer
\([-4,\ 2]\)
The series \(g(x)=\sum_{n=1}^{\infty}\dfrac{(x-2)^n}{\sqrt n\,3^n}\) has radius 3 and interval \([-1,5)\). What is the interval of convergence of the series of \(g'\)?
- \((-1,\ 5)\)
- \([-1,\ 5)\)
- \([-1,\ 5]\)
- \((-1,\ 5]\)
Answer
\((-1,\ 5)\)
Summary
When you differentiate or integrate a power series term by term, the center and the radius stay the same.
The terms at each end change, so an earlier result there may not hold, and you test each end again.
You do this when a problem asks for the interval of convergence of a derivative or integral series.
This is one of five skills in topic 10.15 of Which & Why: AP Calculus.