Handle an inconclusive ratio test

Two series with one ratio limit

The upper graph shows the ratio of neighboring terms for two series. Series P has the terms \(\frac{1}{2n+1}\), and series Q has the terms \(\frac{1}{(2n+1)^2}\).

For both series the dots rise toward the dashed line at 1, so both ratio limits are 1.

The lower graphs show the partial sums. The sums of Q approach 0.234, so Q converges, but the sums of P continue to rise, so P diverges.

The same ratio limit goes with opposite results, so a limit of 1 cannot show whether a series converges.

A limit of 1 gives no conclusion

Drag the slider toward 1. The dots approach the orange line at the height \(L\). When \(L\) is close to 1, each term is almost as large as the one before it, so the terms shrink or grow slowly.

If the terms shrink slowly, the series can converge or it can diverge, and the ratio cannot show how slow is too slow.

Show the two series. The slider moves to L = 1, and the lower graph shows the partial sums of two series with these dots. One levels off, and one continues to rise.

The ratio test cannot tell these two series apart, so you choose another test that uses the terms themselves.

Notation

1. If the ratio limit is 1, the ratio test gives no conclusion about convergence or divergence.

   \[\cy{L} = 1 \ \Rightarrow \ \text{no conclusion}\]

2. Do not write "converges" or "diverges" because of \(L = 1\). Only \(L < 1\) or \(L > 1\) gives a conclusion.

3. Choose a test by the form of the terms. Terms like \(\frac{1}{n^p}\) form a p-series: it converges if \(p > 1\) and diverges if \(p \le 1\). If the terms are smaller than those of a convergent series, or larger than those of a divergent one, use a comparison.

4. If the signs alternate, use the alternating series test. If the terms do not approach 0, use the nth-term test.

   If the terms have powers and roots of \(n\) but no factorial and no \(n\) in an exponent, then \(L = 1\). Use another test from the start.

When it fails

The ratio test uses absolute values, so signs have no effect on it. The series P and the alternating series \(\sum \frac{(-1)^n}{2n+1}\) have the same dots.

The alternating series converges, by the alternating series test, and P diverges. The dots cannot show this difference.

Example 1: Terms like a p-series

Does \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n}}\) converge?

1. Form the ratio and simplify it.

   \[\begin{aligned}\left|\frac{a_{n+1}}{a_n}\right| &= \frac{(n+1)^{-1/3}}{n^{-1/3}}\\ &= \left(\frac{n}{n+1}\right)^{1/3}\end{aligned}\]

2. Divide the top and the bottom of \(\frac{n}{n+1}\) by \(n\). The fraction approaches 1, so \(L = 1\) and the dots approach the dashed line. The ratio test gives no conclusion.

   \[L = \lim_{n\to\infty} \left(\frac{1}{1 + 1/n}\right)^{1/3} = 1\]

3. The terms are \(\frac{1}{n^{1/3}}\), so this is a p-series with \(p = \frac{1}{3}\).

   \[p = \tfrac{1}{3}\]

4. Because \(p \le 1\), the p-series diverges, and its partial sums in the lower graph continue to rise.

Example 2: A ratio that stays below 1

Does \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2+3n}\) converge?

1. Form the ratio, where \((n+1)^2 + 3(n+1) = n^2 + 5n + 4\). Divide the top and the bottom by \(n^2\). The limit is 1, so the ratio test gives no conclusion.

   \[\left|\frac{a_{n+1}}{a_n}\right| = \frac{n^2+3n}{n^2+5n+4} \to 1\]

2. A tempting wrong move: Every ratio is less than 1, so the terms shrink and the series converges.

   The dots lie below the dashed line, but they approach it. The faint dots of series P are also below 1, yet P diverges, so dots below 1 do not show convergence.

3. Compare the terms with the terms of a known series. For \(n \ge 1\), the denominator \(n^2 + 3n\) is larger than \(n^2\), so each term is smaller than \(\frac{1}{n^2}\).

   \[0 < \frac{1}{n^2+3n} < \frac{1}{n^2}\]

4. The series \(\sum \frac{1}{n^2}\) is a p-series with \(p = 2\), so it converges. By comparison, \(\sum \frac{1}{n^2+3n}\) converges, and its partial sums level off.

Practice

Does \(\displaystyle\sum_{n=1}^{\infty} \frac{n+1}{2n+5}\) converge?

Move the slider to large \(n\). The dots approach 1. Then read the value that the bars approach.

Questions

The graph shows the ratio of neighboring terms for two series. Series U has the terms \(\frac{1}{(n+2)^2}\), and series V has the terms \(\frac{1}{n+2}\). The dashed line is at the height 1.

1. What does the ratio test conclude for series U?

   • No conclusion, because the dots approach 1.
   • Convergence, because every dot is below the dashed line.
   • Divergence, because the dots approach 1.
   Answer

   No conclusion, because the dots approach 1.

2. Series U converges, and series V diverges. What does the graph show about the ratio test?

   • The dots of both series approach 1, so the test cannot tell them apart.
   • The dots of V are higher, so the test shows that V diverges.
   • The dots of U are lower, so the test shows that U converges.
   Answer

   The dots of both series approach 1, so the test cannot tell them apart.

3. For which series does the ratio test give no conclusion?

   • \(\displaystyle\sum \frac{2n+1}{n^3+5}\)
   • \(\displaystyle\sum \frac{5^n}{n!}\)
   • \(\displaystyle\sum \frac{n}{3^n}\)
   • \(\displaystyle\sum \frac{n!}{2^n}\)
   Answer

   \(\displaystyle\sum \frac{2n+1}{n^3+5}\)

Summary

When the ratio limit is 1, you write that the ratio test gives no conclusion, and you choose another test that uses the terms.

A convergent series and a divergent series can both have a ratio limit of 1, so the limit cannot tell them apart.

You choose a p-series, a comparison, the alternating series test or the nth-term test by the form of the terms.

This is one of four skills in topic 10.8 of Which & Why: AP Calculus.