Carry out a limit comparison

A ratio that approaches 3/2

The orange bars are the terms \(a_n=\frac{3n^{2}+n}{2n^{4}+5}\). For large \(n\), the leading terms give \(a_n\approx\frac{3n^{2}}{2n^{4}}=\frac32\cdot\frac{1}{n^{2}}\), so the chalk outlines show the terms \(b_n=\frac{1}{n^{2}}\). The lower graph shows the ratio \(\frac{a_n}{b_n}\).

Move n to larger values. The ratio is about 1.55 at n = 10, 1.505 at n = 100 and 1.5005 at n = 1000, so it approaches \(\frac32\).

Turn on the teal strip. From n = 2 on, every dot lies between \(\frac34\) and \(\frac94\). This means every orange bar is between \(\frac34\) and \(\frac94\) times its chalk bar, as the two teal lines in the upper graph show.

The chalk series is a p-series with \(p=2\), so it converges, and \(\frac94\) times it also converges. Every orange bar from n = 2 on is below the upper teal line, so the orange series converges by direct comparison.

A ratio that stays between two positive numbers

Without numbers, the ratio of \(a_n\) to \(b_n\) in the lower graph approaches a number \(c\) that is finite and greater than 0.

The teal strip runs from \(\frac c2\) to \(\frac{3c}{2}\). Because the dots approach \(c\), every dot after some \(n\) lies inside the strip, and the earlier dots are faint. This means each orange bar from there on is between \(\frac c2\) and \(\frac{3c}{2}\) times its chalk bar.

A constant multiple of a series converges or diverges as the series does. If the chalk series converges, the orange series lies below \(\frac{3c}{2}\) times it, so it converges. If the chalk series diverges, the orange series lies above \(\frac c2\) times it, so it diverges.

Notation

1. The terms \(a_n\) and \(b_n\) must both be positive, at least from some \(n\) on.

   \[a_n>0,\qquad b_n>0\]

2. Form the ratio of the two terms, and find its limit.

   \[\lim_{n\to\infty}\frac{\cy{a_n}}{b_n}=c\]

3. If c is finite and greater than 0, the two series behave in the same way. Both converge, or both diverge.

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4. In the opening, the ratio is a fraction of powers. To find its limit, divide the top and the bottom by the highest power of \(n\) in the bottom.

   \[\frac{3n^{4}+n^{3}}{2n^{4}+5}=\frac{3+1/n}{2+5/n^{4}}\to\frac32\]

5. The limit shows whether the series converge. It does not show the sums, which are usually different numbers.

When it fails

Take \(a_n=\frac{2+\sin n}{n^{2}}\) and \(b_n=\frac{1}{n^{2}}\). The ratio is \(2+\sin n\), so the dots move up and down inside the red strip from 1 to 3 and do not approach one number. The limit does not exist, so the test does not apply.

The ratio does stay between 1 and 3, so a direct comparison works: \(\frac{1}{n^{2}}\le a_n\le\frac{3}{n^{2}}\), and the series converges.

Example 1: A divergent comparison

Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{5n+1}{2n^{2}+7}\) converges.

1. Keep the leading terms. They give \(\frac{5n}{2n^{2}}=\frac52\cdot\frac1n\), so choose \(b_n=\frac1n\). Both series have positive terms.

2. Form the ratio, and divide the top and the bottom by \(n^{2}\).

   \[\begin{aligned}\frac{a_n}{b_n}&=\frac{5n^{2}+n}{2n^{2}+7}\\ &=\frac{5+1/n}{2+7/n^{2}}\to\cy{\frac52}\end{aligned}\]

3. The limit is finite and greater than 0. The series of \(\frac1n\) diverges, so the given series diverges.

Example 2: A convergent comparison

Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt{n}+2}{n^{2}+1}\) converges.

1. The leading terms give \(\frac{\sqrt n}{n^{2}}=\frac{1}{n^{3/2}}\), so choose \(b_n=\frac{1}{n^{3/2}}\). Divide the top and the bottom of the ratio by \(n^{2}\).

   \[\begin{aligned}\frac{a_n}{b_n}&=\frac{n^{2}+2n^{3/2}}{n^{2}+1}\\ &=\frac{1+2/\sqrt n}{1+1/n^{2}}\to\cy{1}\end{aligned}\]

2. A tempting wrong move: The limit is 1, so the two series have the same sum.

   \[\sum a_n=\sum b_n\]

   The bottom graph shows the partial sums. After 10 terms, the given series (the teal curve) adds to about 3.35, but the chalk series adds to about 2.00. The limit 1 describes only the terms for large n.

3. The limit 1 is finite and greater than 0, and the p-series with \(p=\frac32\) converges, so the given series converges.

Practice

Decide whether \(\displaystyle\sum_{n=1}^{\infty}\frac{2n+3}{n^{3/2}+1}\) converges. Compare it with \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\).

Divide the top and the bottom of the ratio by \(n^{3/2}\). Move the slider until the dashed line is at the height that the dots approach.

Questions

The dots show the ratio \(\frac{a_n}{b_n}\) of two series with positive terms, for \(n=1\) to 40. The dashed line is at 0.4. The orange strip runs from 0.3 to 0.65.

1. The dots approach 0.4, and the series of \(b_n\) converges. What follows for the series of \(a_n\)?

   • It converges
   • It converges, and its sum is 0.4 times the sum of the series of \(b_n\)
   • It diverges, because the ratio is less than 1
   Answer

   It converges

2. What is the smallest n from which all the dots stay inside the orange strip?

   • n = 4
   • n = 1
   • n = 20
   Answer

   n = 4

3. For \(a_n=\dfrac{4n+9}{n^{3}+2n}\) and \(b_n=\dfrac{1}{n^{2}}\), what is \(\lim\dfrac{a_n}{b_n}\), and what does the test show?

   • The limit is 4, and the series converges
   • The limit is 4, and the series diverges, because 4 is greater than 1
   • The limit is 0, and the series converges
   • The limit is 4, and the sum is 4 times the sum of \(\frac{1}{n^{2}}\)
   Answer

   The limit is 4, and the series converges

Summary

You divide the terms of the two series and find the limit of the ratio.

A finite limit c greater than 0 keeps each term between two positive multiples of the known term, so the series behave alike.

You use this when a direct inequality is hard to prove and the terms look like the terms of a known series.

This is one of four skills in topic 10.6 of Which & Why: AP Calculus.