Check the convergence condition

Six times a power of the ratio

The dots show the partial sums of \(6+6r+6r^2+\cdots\), where the partial sum \(S_n\) is the sum of the first \(n\) terms. At \(r=0.7\), they are 6, 10.2, 13.14 and about 15.2, and they level off at 20. The power \(r^{14}\) in the readout is close to 0.

At \(r=-0.7\), the partial sums swing above and below about 3.53, and each swing is smaller, so they level off there.

At \(r=1.2\), the partial sums grow faster and faster, so they do not level off.

At \(r=1\), every term is 6, so the partial sums 6, 12, 18 and so on grow without bound. At \(r=-1\), they are 6, 0, 6, 0 and so on, so they do not level off either.

The power of the ratio decides

Remove the numbers. The first term is \(a\), and the ratio is \(r\). The dots show the partial sums \(S_n\). The bars below show the powers \(r^n\), green where they are positive and red where they are negative.

If \(r\ne1\), each partial sum is \(\frac{a}{1-r}\) minus \(\frac{a}{1-r}r^n\), as the next section shows. The red segment shows this second part, the gap between a dot and the dashed line.

If \(|r|<1\), the power \(r^n\) approaches 0, so the gap closes and the dots level off at the dashed line.

If \(|r|\ge1\), the power \(r^n\) does not approach 0, so the dots do not level off.

Notation

1. Subtract \(r\) times \(S_n\) from \(S_n\). Every term cancels except the first term \(a\) and the last term \(-ar^n\). If \(r\ne1\), divide by \(1-r\) to find the partial sum.

   \[\begin{aligned}(1-r)\,\cb{S_n}&=a-ar^n\\ \cb{S_n}&=\dfrac{a\left(1-\cy{r^n}\right)}{1-r}\quad(r\ne1)\end{aligned}\]

2. If the absolute value of \(r\) is less than 1, the power \(r^n\) approaches 0, so the partial sums approach \(\frac{a}{1-r}\). A series whose partial sums approach a number converges.

   \[|r|<1\ \Rightarrow\ \lim_{n\to\infty}S_n=\dfrac{a}{1-r}\]

3. If \(|r|\ge1\) and \(a\ne0\), the partial sums do not approach a number, so the series diverges. If \(r\ge1\), they grow in size without bound. If \(r\le-1\), they swing up and down without leveling off.

   The test uses \(|r|\), so the sign of \(r\) does not decide convergence.

When it fails

Set \(r=-1.2\). The terms alternate in sign, as they do for \(r=-0.7\), but now \(|r|=1.2\) is greater than 1.

Each swing of the partial sums is larger than the one before, so the dots do not level off and the series diverges. A negative ratio alone does not make a series converge.

Example 1: A ratio less than 1

Does \(15+6+\frac{12}{5}+\frac{24}{25}+\cdots\) converge?

1. Divide the second term by the first to find the ratio. The third term divided by the second gives the same ratio.

   \[r=\frac{6}{15}=\frac25,\qquad \frac{12/5}{6}=\frac25\]

2. The absolute value of the ratio is \(\frac25\), which is less than 1.

3. The powers of \(\frac25\) approach 0, so the partial sums level off at the dashed line, and the series converges.

Example 2: A ratio that contains x

For which values of \(x\) does \(\displaystyle\sum_{n=0}^{\infty}(2x-1)^n\) converge?

1. Each term is \(2x-1\) times the one before, so the ratio is \(2x-1\). The series converges when its absolute value is less than 1.

2. A tempting wrong move: Solve \(2x-1<1\). This gives \(x<1\).

   \[2x-1<1\ \Rightarrow\ x<1\]

   Take \(x=-1\), which is less than 1. The ratio is \(-3\), and the terms are 1, \(-3\), 9, \(-27\). The partial sums swing wider each time, so the series diverges.

3. The condition is on the absolute value, so the ratio must lie between \(-1\) and 1. Add 1 to all three parts. Then divide all three parts by 2.

   \[-1<2x-1<1\ \Rightarrow\ 0

4. At \(x=0\) the ratio is \(-1\), and at \(x=1\) the ratio is 1. Both series diverge, so the end values are not included.

Practice

The series \(\displaystyle\sum_{n=0}^{\infty}5(\sin x)^n\) has the ratio \(\sin x\). Decide whether it converges for \(x=\frac{\pi}{6}\) and for \(x=\frac{3\pi}{2}\).

Move the slider to the ratio that each value of \(x\) gives. Decide whether the dots level off.

Questions

Questions 1 and 2 use the figure, which shows the partial sums of a geometric series with the first term 8. The dots swing above and below the dashed line at 5.

1. What is the ratio \(r\)?

   • −0.6
   • −4.8
   • 0.4
   Answer

   −0.6

2. Does the series converge? Why?

   • Yes, because \(|r|=0.6\) is less than 1.
   • No, because the partial sums are not always increasing.
   • No, because the ratio is negative.
   Answer

   Yes, because \(|r|=0.6\) is less than 1.

3. What is the ratio of \(\displaystyle\sum_{n=1}^{\infty}\frac{7^n}{5^{n+1}}\), and does the series converge?

   • The ratio is \(\frac75\), and the series diverges.
   • The ratio is \(\frac15\), and the series converges.
   • The ratio is \(\frac75\), and the series converges.
   Answer

   The ratio is \(\frac75\), and the series diverges.

Summary

You test a geometric series by checking whether the absolute value of the ratio is less than 1.

The test works because the partial sums depend on the power \(r^n\), and this power approaches 0 only when \(|r|<1\).

You use the test before you use a sum formula, and for any series whose ratio contains a variable.

This is one of four skills in topic 10.2 of Which & Why: AP Calculus.