10.6: Absolute Convergence and the Ratio and Root Tests

Comparison testing is a powerful tool that excels at assessing the convergence or divergence of rational functions and geometric-like series. But the methods we have discussed so far are difficult to apply to series containing two types of terms—exponential and factorial expressions. This section explores two new tests to cover these weaknesses. We also learn to classify a series' convergence based on the behavior of its positive terms only. Doing so grants us a useful theorem for assessing the convergence of oscillating functions. These ideas are covered by the following topics:

• Absolute and Conditional Convergence
• Ratio Test
• Root Test

Absolute and Conditional Convergence

If a series \(\sum a_n\) converges, then we may wish to classify the convergence either as absolute or conditional. We say \(\sum a_n\) converges absolutely if and only if \(\sum \abs{a_n}\) converges. Conversely, we say \(\sum a_n\) converges conditionally if and only if \(\sum a_n\) converges but \(\sum \abs{a_n}\) diverges.

If \(\sum a_n\) converges absolutely, then \(\sum a_n\) converges. We may use this fact as a convergence test; that is, if we can establish that \(\sum \abs{a_n}\) converges, then we can assert that \(\sum a_n\) also converges. We use this pathway if it is easier to test the convergence or divergence of \(\sum \abs{a_n}.\)

PROOF Since \(\abs{a_n}\) is either \(a_n\) or \(-a_n,\) we have \[0 \leq a_n + \abs{a_n} \leq 2 \abs{a_n} \pd\] Assuming \(\sum \abs{a_n}\) converges, the series \(2\sum \abs{a_n}\) is also convergent. Thus, by the Direct Comparison Test, \(\sum \par{a_n + \abs{a_n}}\) is convergent. But \[\sum a_n = \sum \par{a_n + \abs{a_n}} - \sum \abs{a_n}\] is a difference of two convergent series. Since the sum or difference of two convergent series is a convergent series, \(\sum a_n\) converges. \[\qedproof\]

Ratio Test

Consider the infinite series \(\sum a_n.\) The Ratio Test enables us to determine whether the series converges by considering the ratio \(\abs{a_{n + 1}/a_n}\) as \(n\) grows large. The test is as follows.

This test is particularly useful when \(a_n\) contains factorial or exponential terms. The following property is applicable when we decompose factorials: \begin{equation} (n + k)! = (n + k) (n + k - 1) \cdot \, \cdots \, \cdot (n + 1) \, n! \cma \label{eq:factorial-decompose} \end{equation} where \(k \geq 1\) is a positive integer. For example, \[(n + 3)! = (n + 3)(n + 2)(n + 1) \, n! \pd\] You may confirm this fact with integers; for example, \(5! = (5)(4)(3!).\)

From Example 5, we conclude that alternators in the form \((-1)^n\) or \((-1)^{n + 1}\) (whose absolute values are \(1\)) have no effect on the quantity \(\abs{a_{n + 1}/a_n}.\) Thus, we can ignore these alternators when we set up ratios.

PROOF OF THE RATIO TEST Consider the infinite series \(\sum_{n = 1}^\infty a_n.\) We begin the index at \(n = 1\) with no loss of generality; that is, this proof applies for other starting indices as well. Let \[L = \lim_{n \to \infty} \abs{\frac{a_{n + 1}}{a_n}} \pd \] We aim to show that \(\sum_{n = 1}^\infty a_n\) converges absolutely if \(L \lt 1\) and that \(\sum_{n = 1}^\infty a_n\) diverges if \(L \gt 1.\) For \(L = 1\) we will select values of \(a_n\) such that \(\sum_{n = 1}^\infty a_n\) may either converge absolutely, converge conditionally, or diverge.

Case 1 We want to show that \(\sum_{n = 1}^\infty a_n\) converges absolutely—that is, \(\sum_{n = 1}^\infty \abs{a_n}\) converges—if \(L \lt 1.\) To do so, we compare \(\sum_{n = 1}^\infty \abs{a_n}\) to an infinite geometric series. Let \(r\) be a number such that \(L \lt r \lt 1.\) Then eventually, as \(n\) grows large, the ratio \(\abs{a_{n + 1}/a_n}\) becomes bounded by \(r.\) For some integer \(N,\) if \(n \geq N,\) then we have \[\abs{\frac{a_{n + 1}}{a_n}} \lt r \implies \abs{a_{n + 1}} \lt r \abs{a_n} \pd\] We consider the following: \[ \baat{2} \abs{a_{N + 1}} & \lt r \abs{a_{N}} \nl \abs{a_{N + 2}} & \lt r \abs{a_{N + 1}} && \lt r^2 \abs{a_N} \nl \abs{a_{N + 3}} & \lt r \abs{a_{N + 2}} && \lt r^3 \abs{a_N} \nl & \; \; \vdots \nl \abs{a_{N + k}} & \lt r \abs{a_{N + k - 1}} && \lt r^k \abs{a_N} \pd \nl \eaat \] Note that \(\sum_{k = 1}^\infty \abs{a_N} r^k\) is a convergent geometric series since \(0 \lt r \lt 1.\) Because \(\abs{a_{N + k}} \lt r^k \abs{a_N},\) using the Direct Comparison test shows that \[\sum_{k = 1}^\infty \abs{a_{N + k}} = \sum_{n = N+1}^\infty \abs{a_n}\] converges. But \[\sum_{n = 1}^\infty \abs{a_n} = \sum_{n = 1}^N \abs{a_n} + \sum_{n = N + 1}^\infty \abs{a_n} \pd\] On the right-hand side, \(\sum_{n = 1}^N \abs{a_n}\) is a sum of finite terms, and we established that \(\sum_{n = N + 1}^\infty \abs{a_n}\) is convergent. Adding a finite number to a convergent series yields another convergent series. Thus, \(\sum_{n = 1}^\infty \abs{a_n}\) converges and so \(\sum_{n = 1}^\infty a_n\) converges absolutely if \(L \lt 1.\)

Case 2 We must now show that if \(L \gt 1,\) then \(\sum_{n = 1}^\infty a_n\) diverges. Because \(L \gt 1,\) there exists an integer \(N\) such that for \(n \geq N,\) \[\abs{\frac{a_{n + 1}}{a_n}} \gt 1 \implies \abs{a_{n + 1}} \gt \abs{a_n} \pd\] This inequality asserts that the terms of \(\abs{a_n}\) are growing and never negative. Consequently, \(\lim_{n \to \infty} a_n \ne 0,\) so by the Divergence Test \(\sum_{n = 1}^\infty a_n\) diverges.

Case 3 To show that the Ratio Test is inconclusive for \(L = 1,\) we need to prove that no consistent result exists. We may brainstorm series \(\sum_{n = 1}^\infty a_n\) that satisfy \[L = \lim_{n \to \infty} \abs{\frac{a_{n + 1}}{a_n}} = 1 \pd\] For example, \(\sum_{n = 1}^\infty (1/n)\) is divergent and satisfies \(L = 1.\) But \(\sum_{n = 1}^\infty (-1)^n/n\) is conditionally convergent while also satisfying \(L = 1.\) And \(\sum_{n = 1}^\infty (1/n^2)\) is an absolutely convergent series such that \(L = 1.\) Thus, the Ratio Test is inconclusive for \(L = 1.\) \[\qedproof\]

Root Test

Consider the infinite series \(\sum a_n.\) If the summation function \(a_n\) contains a power of \(n,\) then we may find value in using the Root Test, as follows.

Note the similarity of the conclusions in the Root Test and the Ratio Test. In many cases, using either test yields the same conclusion about a series' convergence or divergence.

When we use the Root Test, we often need to take the \(n\)th root of the term \(n^k,\) where \(k \gt 0.\) In these cases, we need to find the value of \[\lim_{n \to \infty} \par{n^k}^{1/n} = \lim_{n \to \infty} n^{k/n} \pd \] Evaluating this indeterminate limit using the methods of Section 3.5, we obtain \begin{equation} \lim_{n \to \infty} n^{k/n} = 1 \pd \label{eq:lim-power} \end{equation}

PROOF OF THE ROOT TEST This proof is similar to the Proof of the Ratio Test. We start with the infinite series \(\sum_{n = 1}^\infty a_n,\) where we begin the index at \(n = 1\) with no loss of generality, just as we did when proving the Ratio Test. Let \[L = \lim_{n \to \infty} \sqrt[n]{\abs{a_n}} = \lim_{n \to \infty} \abs{a_n}^{1/n} \pd\] Our goal is to show that \(\sum_{n = 1}^\infty a_n\) converges absolutely if \(L \lt 1\) and that \(\sum_{n = 1}^\infty a_n\) diverges if \(L \gt 1.\) If \(L = 1,\) then we select values of \(a_n\) such that \(\sum_{n = 1}^\infty a_n\) may either converge absolutely, converge conditionally, or diverge—just as we did when we proved the Ratio Test.

Case 1 If \(L \lt 1,\) then we want to show that \(\sum_{n = 1}^\infty a_n\) is absolutely convergent. We define a number \(r\) such that \(L \lt r \lt 1.\) Because \(\abs{a_n}^{1/n} \to L,\) there exists a number \(N\) such that for \(n \geq N,\) \[\abs{a_n}^{1/n} \lt r \implies \abs{a_n} \lt r^n \pd\] The series \(\sum_{n = 1}^\infty r^n\) is an infinite geometric series that converges because \(0 \lt r \lt 1.\) Therefore, by the Direct Comparison Test the series \(\sum_{n = N}^\infty \abs{a_n}\) is also convergent. But \[\sum_{n = 1}^\infty \abs{a_n} = \sum_{n = 1}^{N - 1} \abs{a_n} + \sum_{n = N}^\infty \abs{a_n} \pd\] On the right-hand side the term \(\sum_{n = 1}^{N - 1} \abs{a_n}\) is a finite sum, and we established that \(\sum_{n = N}^\infty \abs{a_n}\) is convergent. Adding a finite number to a convergent series yields a convergent series. Hence, \(\sum_{n = 1}^\infty \abs{a_n}\) converges and so \(\sum_{n = 1}^\infty a_n\) converges absolutely.

Case 2 We now show that \(\sum_{n = 1}^\infty a_n\) is divergent if \(L \gt 1.\) Since \(L \gt 1,\) we know that \(\abs{a_n}^{1/n}\) approaches a number greater than \(1\) as \(n\) grows large—say, for \(n \geq N \col\) \[\abs{a_n}^{1/n} \gt 1 \implies \abs{a_n} \gt 1^n = 1 \pd\] Because \(\abs{a_n} \gt 1\) for all \(n \geq N,\) we assert that \(\lim_{n \to \infty} a_n \ne 0.\) Thus, the Divergence Test states that \(\sum_{n = 1}^\infty a_n\) diverges.

Case 3 To show that the Root Test is inconclusive for \(L = 1,\) we simply show that no consistent result exists. We consider series \(\sum_{n = 1}^\infty a_n\) that satisfy \[L = \lim_{n \to \infty} \abs{a_n}^{1/n} = 1 \pd\] For example, \(\sum_{n = 1}^\infty (1/n)\) is divergent and satisfies \(L = 1.\) Yet \(\sum_{n = 1}^\infty (-1)^n/n\) is conditionally convergent while also satisfying \(L = 1,\) and \(\sum_{n = 1}^\infty (1/n^2)\) is an absolutely convergent series such that \(L = 1.\) Thus, the Root Test is inconclusive for \(L = 1.\) \[\qedproof\]