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10.4: Comparison Tests
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10.4: Comparison Tests

We have discussed how to test \(p\)-series and geometric series for convergence or divergence. In Section 10.3 we learned to test a series for convergence or divergence by computing an improper integral. But some series are very similar to other series known to converge or diverge. For example, the series \(\sum_{n = 1}^\infty \par{1/3^n}\) is geometric, whereas \(\sum_{n = 1}^\infty \parbr{1/\par{3^n + 1}}\) is nearly identical but not geometric. Likewise, \(\sum_{n = 1}^\infty \par{1/n^4}\) is a \(p\)-series, while \(\sum_{n = 1}^\infty \parbr{1/\par{n^4 + 1}}\) is similar but isn't a \(p\)-series. These examples motivate Comparison Tests, in which we assert that a series converges or diverges by comparing it to another series known to be convergent or divergent. We discuss the following tests:

Direct Comparison Test

Let \(\{a_n\}\) and \(\{b_n\}\) be sequences whose terms are all positive. Suppose that the terms of a sequence \(\{a_n\}\) are always less than or equal to the terms of the sequence \(\{b_n\}.\) Then the partial sum \(\sum_{i = 1}^N a_i\) is less than or equal to the partial sum \(\sum_{i = 1}^N b_i.\) Mathematically, we write \[0 \leq \sum_{i = 1}^N a_i \leq \sum_{i = 1}^N b_i\] since in the middle partial sum we add up smaller numbers. But what happens as \(N \to \infty \ques\) If \(\lim_{N \to \infty} \sum_{i = 1}^N b_i\) exists and equals \(T,\) then the inequality becomes \[0 \leq \lim_{N \to \infty} \sum_{i = 1}^N a_i \leq T \or 0 \leq \sum_{n = 1}^\infty a_n \leq T \pd\] Accordingly, \(\sum_{n = 1}^\infty a_n\) is bounded between two finite numbers and so converges. Therefore, if \(\sum_{n = 1}^\infty b_n\) converges and \(0 \lt a_n \leq b_n,\) then \(\sum_{n = 1}^\infty a_n\) also converges. Informally, if the larger series converges, then the smaller series also converges.

Conversely, let's suppose that the terms of \(\{a_n\}\) are always greater than or equal to those of \(\{b_n\}.\) If \(a_n \gt 0\) and \(b_n \gt 0,\) then the \(N\)th partial sums of \(a_i\) and \(b_i\) are related by \[\sum_{i = 1}^N a_i \geq \sum_{i = 1}^N b_i \gt 0\] because \(a_n \geq b_n \gt 0.\) If \(\sum_{n = 1}^\infty b_n\) diverges, then the partial sum \(\sum_{i = 1}^N b_i\) approaches infinity as \(N \to \infty\) since \(b_n \gt 0.\) (A series can diverge due to either oscillation or becoming unbounded. But since \(b_n \gt 0,\) the sum keeps increasing and therefore cannot be oscillatory.) Accordingly, \(\sum_{n = 1}^\infty a_n\) must also diverge since it is greater than the divergent series \(\sum_{n = 1}^\infty b_n.\) So if the smaller series diverges, then the larger series also diverges. We summarize these ideas as the Direct Comparison Test.

DIRECT COMPARISON TEST
  1. If \(0 \leq a_n \leq b_n\) and \(\sum b_n\) converges, then \(\sum a_n\) also converges.
  2. If \(a_n \geq b_n \geq 0\) and \(\sum b_n\) diverges, then \(\sum a_n\) also diverges.

PROOF OF DIRECT COMPARISON TEST

  1. Consider the three sums \[\andThree {S_N = \sum_{i = 1}^N a_i} {T_N = \sum_{i = 1}^N b_i} {T = \sum_{n = 1}^\infty b_n} \cma \] where \(\{S_N\}\) and \(\{T_N\}\) have positive terms and \(\sum_{n = 1}^\infty b_n\) is convergent. If \(0 \lt a_i \leq b_i,\) then \(S_N \leq T_N.\) As \(N \to \infty,\) \(T_N \to T.\) Because \(b_i \gt 0,\) the sequence \(T_N\) is increasing and bounded above, so \(T_N \leq T\) for all \(N.\) Therefore, \(S_N \leq T,\) meaning \(\{S_N\}\) is bounded above and therefore converges. Since \[\lim_{N \to \infty} S_N = \lim_{N \to \infty} \sum_{i = 1}^N a_i = \sum_{n = 1}^\infty a_n \cma \] the series \(\sum_{n = 1}^\infty a_n\) converges.
  2. Suppose that \(\sum_{n = 1}^\infty b_n\) diverges, and let \[S_N = \sum_{i = 1}^N a_i \and T_N = \sum_{i = 1}^N b_i \pd\] If \(a_i \geq b_i \gt 0,\) then \(S_N \geq T_N \gt 0.\) Since \(\sum_{n = 1}^\infty b_n\) diverges, \(T_N \to \infty\) as \(N \to \infty.\) But because \(S_N\) is greater than or equal to \(T_N,\) it follows that \(S_N \to \infty\) too. Since \[\lim_{N \to \infty} S_N = \lim_{N \to \infty} \sum_{i = 1}^N a_i = \sum_{n = 1}^\infty a_n \cma\] the series \(\sum_{n = 1}^\infty a_n\) also diverges.
\[\qedproof\]

The Direct Comparison Test is useful only if we can readily identify a series \(\sum b_n\) that is known to converge or diverge. In testing \(\sum a_n\) for convergence or divergence, our goal is to either find a smaller divergent sum \(\sum b_n\) or a larger convergent sum \(\sum b_n.\) In many cases, we perform comparisons using the following choices of \(\sum b_n \col\)

EXAMPLE 1
Determine whether the following series converges or diverges: \[\sum_{n = 1}^\infty \frac{2}{n^2 + n + 4} \pd\]
For \(n \geq 1\) we see \(n^2 + n + 4 \geq n^2,\) and both expressions are positive. Accordingly, \[0 \lt \frac{1}{n^2 + n + 4} \leq \frac{1}{n^2} \and 0 \lt \frac{2}{n^2 + n + 4} \leq \frac{2}{n^2} \pd\] Thus, we compare \(\sum_{n = 1}^\infty \parbr{2/\par{n^2 + n + 4}}\) to \[\sum_{n = 1}^\infty b_n = \sum_{n = 1}^\infty \frac{2}{n^2} \pd\] We identify \(\sum_{n = 1}^\infty \par{1/n^2}\) as a \(p\)-series with \(p = 2 \gt 1,\) so it converges. Multiplying a convergent series by a constant gives another convergent series; thus, \(\sum_{n = 1}^\infty \par{2/n^2}\) \(= \sum_{n = 1}^\infty b_n\) converges. Part (a) of the Direct Comparison Test therefore asserts that \[\sum_{n = 1}^\infty \frac{2}{n^2 + n + 4} \convBoxed\]
EXAMPLE 2
Determine whether the following series converges or diverges: \[\sum_{n = 2}^\infty \frac{1}{\sqrt n - 1} \pd\]
We want to find either a convergent series that's larger than \(\sum_{n = 2}^\infty \parbr{1/\par{\sqrt n - 1}}\) or a divergent series that's smaller than \(\sum_{n = 2}^\infty \parbr{1/\par{\sqrt n - 1}}.\) For \(n \geq 2\) we note that \(\sqrt n \geq \sqrt n - 1.\) Thus, \[ \frac{1}{\sqrt{n} - 1} \geq \frac{1}{\sqrt n} \gt 0 \pd \] We therefore compare \(\sum_{n = 2}^\infty \parbr{1/\par{\sqrt n - 1}}\) to \[\sum_{n = 2}^\infty b_n = \sum_{n = 2}^\infty \frac{1}{\sqrt n} \cma\] which diverges because it is a \(p\)-series with \(p = 1/2 \leq 1.\) Hence, part (b) of the Direct Comparison Test asserts that \[\sum_{n = 2}^\infty \frac{1}{\sqrt n - 1} \divBoxed\]

REMARK The inequalities required by the Direct Comparison Test only need to be true for \(n\) after a certain point. In other words, either inequality \(0 \lt a_n \leq b_n\) or \(a_n \geq b_n \gt 0\) needs to be satisfied for \(n \geq N,\) where \(N\) is a constant integer that doesn't need to match the sum's starting index. This is because adding a finite number of terms doesn't alter a series' convergence, as shown by the following example.

EXAMPLE 3
Determine whether the following series converges or diverges: \[\sum_{n = 1}^\infty \frac{\ln n}{\sqrt[3]{n}} \pd\]
Our goal is to identify either a convergent series that's larger than \(\sum_{n = 1}^\infty (\ln n)/\sqrt[3]{n}\) or a divergent series that's smaller than \(\sum_{n = 1}^\infty (\ln n)/\sqrt[3]{n}.\) Because \(\ln n \geq 1\) for \(n \geq 3,\) it holds that \[\frac{\ln n}{\sqrt[3]{n}} \geq \frac{1}{\sqrt[3]{n}} \gt 0 \cmaa n \geq 3 \pd\] We therefore compare \(\sum_{n = 3}^\infty (\ln n)/\sqrt[3]{n}\) to \[\sum_{n = 3}^\infty b_n = \sum_{n = 3}^\infty \frac{1}{\sqrt[3]{n}} \cma\] which diverges because it is a \(p\)-series with \(p = 1/3 \leq 1.\) Thus, part (b) of the Direct Comparison Test states that \[\sum_{n = 3}^\infty \frac{\ln n}{\sqrt[3]{n}}\ \div \pd\] Because \[\sum_{n = 1}^\infty \frac{\ln n}{\sqrt[3]{n}} = \sum_{n = 1}^2 \frac{\ln n}{\sqrt[3]{n}} + \sum_{n = 3}^\infty \frac{\ln n}{\sqrt[3]{n}}\] and \(\sum_{n = 1}^2 (\ln n)/\sqrt[3]{n}\) is finite, the infinite series \[\sum_{n = 1}^\infty \frac{\ln n}{\sqrt[3]{n}}\ \divBoxed\] because adding a finite number to a divergent series yields another divergent series.

In testing the series \(\sum a_n\) for convergence (where \(a_n\) is positive), the Direct Comparison Test fails if we cannot find a convergent series \(\sum b_n\) that's larger than \(\sum a_n\) or if we cannot find a divergent series \(\sum b_n\) that's smaller than \(\sum a_n.\) For example, it would be convenient to compare the series \[\sum_{n = 1}^\infty a_n = \sum_{n = 1}^\infty \frac{1}{3^n - 1}\] to \(\sum_{n = 1}^\infty b_n\) \(= \sum_{n = 1}^\infty \par{1/3^n},\) a convergent geometric series of common ratio \(r = 1/3 \lt 1.\) But for \(n \geq 1,\) we see \[\frac{1}{3^n - 1} \geq \frac{1}{3^n} \gt 0 \pd\] This inequality implies that \(\sum_{n = 1}^\infty \parbr{1/\par{3^n - 1}}\) could either converge or diverge because it is greater than the convergent series \(\sum_{n = 1}^\infty \par{1/3^n}.\) And so the Direct Comparison Test is useless. Yet astonishingly, the Direct Comparison Test holds that the similar series \(\sum_{n = 1}^\infty \parbr{1/\par{3^n + 1}}\) converges because for \(n \geq 1,\) \[0 \lt \frac{1}{3^n + 1} \leq \frac{1}{3^n} \cma\] so \(\sum_{n = 1}^\infty \parbr{1/\par{3^n + 1}}\) is bounded between two finite numbers. Regardless, our intuition wants to conclude that \(\sum_{n = 1}^\infty \parbr{1/\par{3^n - 1}}\) converges because it's nearly identical to \(\sum_{n = 1}^\infty \parbr{1/\par{3^n + 1}}.\) This guess turns out to be correct, which we prove using the Limit Comparison Test.

Limit Comparison Test

When we cannot establish an inequality required to use the Direct Comparison Test, we resort to another test—the Limit Comparison Test. If \(a_n\) and \(b_n\) are both positive, then we consider the ratio \(a_n/b_n\) as \(n \to \infty.\) Let \[L = \lim_{n \to \infty} \frac{a_n}{b_n} \pd\] Recall the idea of relative growth rates from Section 3.5: If \(L\) is a positive, finite number, then \(a_n\) and \(b_n\) grow at the same rate as \(n\) is increased. This relationship between \(\sum a_n\) and \(\sum b_n\) suggests that they either both converge or both diverge. The test is formally stated as follows.

LIMIT COMPARISON TEST
Suppose that \(a_n\) and \(b_n\) are both positive, and let \[L = \lim_{n \to \infty} \frac{a_n}{b_n} \pd\] If \(L\) is positive and finite, then the Limit Comparison Test states that \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge.

TIP If \(a_n/b_n\) approaches a positive, finite number as \(n \to \infty,\) then the ratio \(b_n/a_n\) also approaches a positive, finite number as \(n \to \infty.\) Accordingly, the way by which we divide the summation functions \(a_n\) and \(b_n\) does not matter.

PROOF OF THE LIMIT COMPARISON TEST Let \(m\) and \(M\) be positive numbers such that \(m \lt L \lt M,\) where \(a_n/b_n\) is close to \(L\) for large \(n\)—say, for \(n \gt N.\) We therefore have, for \(n \gt N,\) \[m \lt \frac{a_n}{b_n} \lt M \or m b_n \lt a_n \lt M b_n \pd\] If \(\sum b_n\) converges, then so do \(\sum mb_n\) and \(\sum Mb_n.\) Part (a) of the Direct Comparison Test therefore asserts that \(\sum a_n\) converges, as it is bounded between two convergent sums. Conversely, if \(\sum b_n\) diverges, then \(\sum mb_n\) and \(\sum Mb_n\) both diverge as well. Because \(a_n \gt m b_n \geq 0\) and \(\sum m b_n\) diverges, part (b) of the Direct Comparison Test forces \(\sum a_n\) to also diverge. \[\qedproof\]

EXAMPLE 4
Test the following series for convergence or divergence: \[\sum_{n = 1}^\infty \frac{1}{3^n - 1} \pd\]
Let \(a_n = 1/\par{3^n - 1}.\) We compare \(\sum_{n = 1}^\infty a_n\) to \[\sum_{n = 1}^\infty b_n = \sum_{n = 1}^\infty \frac{1}{3^n} \cma\] which converges because it is a geometric series of common ratio \(r = 1/3\) \(\lt 1.\) Then we see \[ \ba L &= \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\dfrac{1}{3^n - 1}}{1/3^n} \nl &= \lim_{n \to \infty} \frac{3^n}{3^n - 1} = \lim_{n \to \infty} \frac{1}{1 - 1/3^n} \nl &= \frac{1}{1 - 0} = 1 \pd \ea \] Since \(L\) is positive and finite, the Limit Comparison Test asserts that \[\sum_{n = 1}^\infty \frac{1}{3^n - 1} \convBoxed\] as we suspected due to its similarity to \(\sum_{n = 1}^\infty \par{1/3^n}.\)

Sums of Algebraic Functions We often test for convergence or divergence a series \(\sum a_n\) where \(a_n\) is a fraction containing powers or radicals. To do so, we compare \(\sum a_n\) to an appropriate sum \(\sum b_n\) by keeping the highest powers in the numerator and denominator, as shown by the following examples. Computing limits at infinity is therefore crucial; to review this concept, see Section 1.6.

EXAMPLE 5
Test the following series for convergence or divergence: \[\sum_{n = 1}^\infty \frac{n^2 + 5n - 3}{n^3 + 2} \pd\]
Considering only the highest powers in the numerator and denominator, we compare this series to \[\sum_{n = 1}^\infty \frac{n^2}{n^3} = \sum_{n = 1}^\infty \frac{1}{n} \cma\] the divergent Harmonic series. Hence, we choose \(b_n = 1/n.\) Dividing both summation functions, we see \[ \ba L &= \lim_{n \to \infty} \frac{\dfrac{n^2 + 5n - 3}{n^3 + 2}}{1/n} \nl &= \lim_{n \to \infty} \frac{n^3 + 5n^2 - 3n}{n^3 + 2} \nl &= 1 \pd \ea \] Because \(L\) is finite and positive, the Limit Comparison Test says that \[\sum_{n = 1}^\infty \frac{n^2 + 5n - 3}{n^3 + 2} \divBoxed\]
EXAMPLE 6
Test the following series for convergence or divergence: \[\sum_{n = 2}^\infty \frac{\sqrt{n^2 + n + 4}}{3n^4 - 5n + 2} \pd\]
We compare this series to another series with only the highest powers in the numerator and denominator: \[\sum_{n = 2}^\infty \frac{\sqrt{n^2}}{3n^4} = \sum_{n = 2}^\infty \frac{\abs n}{3n^4} = \sum_{n = 2}^\infty \frac{1}{3n^3} \pd\] This series converges because it is a \(p\)-series with \(p = 3 \gt 1.\) Thus, we let \(b_n = 1/(3n^3)\) and consider the limit \[ \ba L &= \lim_{n \to \infty} \frac{\dfrac{\sqrt{n^2 + n + 4}}{3n^4 - 5n + 2}}{1/(3n^3)} \nl &= \lim_{n \to \infty} \frac{\sqrt{n^2 + n + 4}}{3n^4 - 5n + 2} \cdot 3n^3 \nl &= \lim_{n \to \infty} \frac{3 \sqrt{n^2 + n + 4}}{3n - 5/n^2 + 2/n^3} \pd \ea \] In the numerator, factoring out \(\sqrt{n^2} = \abs n = n\) (since \(n \gt 0\)) gives \[ \ba L &= \lim_{n \to \infty} \frac{3n \sqrt{1 + 1/n + 4/n^2}}{3n - 5/n^2 + 2/n^3} \nl &= \lim_{n \to \infty} \frac{3 \sqrt{1 + 1/n + 4/n^2}}{3 - 5/n^3 + 2/n^4} \nl &= \frac{3 \sqrt{1 + 0 + 0}}{3 - 0 + 0} \nl &= 1 \pd \ea \] (The same result can also be attained by using L'Hôpital's Rule.) Because \(L\) is positive and finite, the Limit Comparison Test says that \[\sum_{n = 2}^\infty \frac{\sqrt{n^2 + n + 4}}{3n^4 - 5n + 2} \convBoxed\]

Direct Comparison Test The Direct Comparison Test enables us to establish whether a positive sum \(\sum a_n\) converges or diverges by comparing \(\sum a_n\) to either a larger convergent sum \(\sum b_n\) or a smaller divergent sum \(\sum b_n.\)

  1. If \(0 \leq a_n \leq b_n\) and \(\sum b_n\) converges, then \(\sum a_n\) also converges.
  2. If \(a_n \geq b_n \geq 0\) and \(\sum b_n\) diverges, then \(\sum a_n\) also diverges.

Limit Comparison Test Suppose that \(a_n\) and \(b_n\) are both positive, and let \[L = \lim_{n \to \infty} \frac{a_n}{b_n} \pd\] If \(L\) is positive and finite, then the Limit Comparison Test states that \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge. If \(a_n\) is a fraction with powers and radicals, then we compare \(\sum a_n\) to an appropriate sum \(\sum b_n\) by keeping the highest powers in the numerator and denominator of \(a_n.\)