Check transformed inputs
The input is 3x, not x
The series \(\sum_{n=0}^{\infty}(3x)^n\) is the geometric series \(\sum u^n\) with \(u=3x\), and the geometric series converges for \(u\) between −1 and 1.
The top line shows \(x\), and the bottom line shows \(u\). Drag the orange point on the top line. The point on the bottom line is at \(u=3x\), so it moves three times as fast.
The series converges while the bottom point is inside the teal band from −1 to 1. Because \(3x=1\) at \(x=\frac13\), this happens for \(x\) between \(-\frac13\) and \(\frac13\). Press the button to show this band on the top line.
The band for \(x\) has length \(\frac23\), a third of the band for \(u\), because \(u=3x\) triples every distance.
The condition moves with the input
The teal band on the bottom line shows the inputs u for which a known series converges. A rule u = g(x) gives u from x, and a thin line joins x to its u.
Choose a rule with the buttons. Then drag the point x. The teal band on the top line holds the x for which g(x) lies in the bottom band.
The new series converges at x when g(x) is inside the bottom band, and only then.
Notation
Start with a series in \(u\) that converges for \(|u|\lt R\).
\[\sum c_n\,u^n,\qquad |u|\lt R\]The new series has \(u=g(x)\). Put \(g(x)\) in place of \(u\) in the condition.
\[|g(x)|\lt R\]Solve the inequality for x. If you multiply or divide by a negative number, reverse the inequalities.
\[\begin{aligned}&|kx|\lt R\iff |x|\lt \frac{R}{|k|} \\ &|x-h|\lt R \\ &\quad\iff h-R\lt x\lt h+R\end{aligned}\]If \(g(x)\) equals \(R\) or \(-R\), the new series at that \(x\) is the known series at an end. It converges there when the known series converges at that end.
When it fails
The rule \(u=\frac1x\) is not a scale or a shift. With the geometric series, the condition is \(\left|\frac1x\right|\lt 1\).
This holds when \(|x|>1\), so the teal region on the top line is two pieces and the series diverges near 0. Solve the inequality for \(x\), because the \(x\) that work need not form one interval.
Example 1: Scale and shift
Find where \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{x-2}{3}\right)^n\) converges.
Let \(u=\frac{x-2}{3}\), so the series is \(\sum\frac{u^n}{n}\). The ratio test gives \(|u|\lt 1\). At \(u=1\) the series is harmonic and diverges, and at \(u=-1\) it alternates and converges.
\[-1\le u\lt 1\]Put \(u=\frac{x-2}{3}\) into the condition.
\[-1\le\frac{x-2}{3}\lt 1\]Multiply every part by 3. Then add 2 to every part.
\[-3\le x-2\lt 3\ \Rightarrow\ -1\le x\lt 5\]The end \(x=-1\) gives \(u=-1\), so the series converges there. The end \(x=5\) gives \(u=1\), so it diverges there.
\[[-1,\ 5)\]
Example 2: A negative scale
Find where \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\left(-\frac{x}{2}\right)^n\) converges.
Let \(u=-\frac{x}{2}\). The series \(\sum\frac{u^n}{n}\) converges for \(-1\le u\lt 1\), as in Example 1.
\[-1\le-\frac{x}{2}\lt 1\]A tempting wrong move: Multiply every part by −2 and keep the signs of the inequalities. This gives \(-2\le x\lt 2\).
\[-2\le x\lt 2\]At \(x=2\), the value \(u\) is −1, and the series converges. At \(x=-2\), the value \(u\) is 1, and the series is harmonic, so it diverges. This answer includes −2 and leaves out 2.
Multiplying by a negative number reverses each inequality. Multiply every part by −2 and reverse both inequality signs.
\[2\ge x>-2\]The series converges for \(x\) from −2, not included, to 2, included.
\[(-2,\ 2]\]
Practice
Find where \(\displaystyle\sum_{n=1}^{\infty}\frac{(2x-1)^n}{n^2}\) converges.
Let \(u=2x-1\), and drag \(x\) until the bottom point reaches each end of the band. At \(u=\pm1\), the terms have absolute value \(\frac{1}{n^2}\).
Solution
Let \(u=2x-1\). The ratio test gives \(|u|\lt 1\). At \(u=\pm1\), the terms have absolute value \(\frac{1}{n^2}\), and \(\sum\frac{1}{n^2}\) converges, so the series converges for \(-1\le u\le1\).
\[-1\le 2x-1\le 1\]Add 1 to every part. Then divide every part by 2.
\[0\le 2x\le 2\ \Rightarrow\ 0\le x\le 1\]The ends \(x=0\) and \(x=1\) give \(u=-1\) and \(u=1\), where the series converges, so the interval is closed.
\[[0,\ 1]\]
Questions
The teal band on the bottom line is the range of \(u\) where a known series converges. The top line shows \(x\), and \(u=\frac{x+1}{2}\).
What does the figure show about the series at x = 2?
- It diverges, because u = 1.5 lies outside the band.
- It converges, because x = 2 lies in the band.
- It converges, because u is close to 1.
Answer
It diverges, because u = 1.5 lies outside the band.
Which range of x does the teal band on the top line show?
- −3 < x < 1
- −1 < x < 1
- −2 < x < 0
- −1 < x < 3
Answer
−3 < x < 1
Where does \(\displaystyle\sum_{n=0}^{\infty}\left(\frac{4-3x}{5}\right)^n\) converge?
- \(-\frac13\lt x\lt 3\)
- \(-\frac53\lt x\lt \frac53\)
- \(1\lt x\lt \frac53\)
- \(-3\lt x\lt \frac13\)
Answer
\(-\frac13\lt x\lt 3\)
Summary
A new series converges at x only when its input g(x) lies where the known series converges.
You solve |g(x)| < R for x and then test each end with the known series.
This finds where any series built from a known series by a scale, a shift or a substitution converges.
This is one of four skills in topic 10.13 of Which & Why: AP Calculus.