Why Check the Absolute Values Separately?
An alternating sign can make a series converge even when the series of absolute values diverges. For rational terms, the degrees of the numerator and denominator give a useful first prediction: they reveal the power of \(n\) that the absolute values resemble. But that prediction answers only the absolute-convergence question. To conclude that the original alternating series converges, you must also check the conditions of the Alternating Series Test.
The method is therefore a two-part analysis. First check whether the magnitudes decrease eventually and tend to zero. Then discard the signs and compare the absolute-value series with an appropriate \(p\)-series. These checks lead to three possible conclusions: absolute convergence, conditional convergence, or divergence.
Rational Terms and Their Leading Powers
Consider a rational sequence \(b_n=P(n)/Q(n)\), where \(P\) and \(Q\) are polynomials and \(b_n\) is positive for all sufficiently large \(n\). Let \(d_P\) and \(d_Q\) be their degrees, and let \(c_P\) and \(c_Q\) be their leading coefficients. For large \(n\), the leading terms determine the ratio:
Here, \(b_n\sim c n^{-p}\) means that \(b_n/(c n^{-p})\) tends to \(1\) as \(n\) tends to infinity. When \(d_Q>d_P\), set \(p=d_Q-d_P\). Then \(b_n\) behaves like a positive constant times \(1/n^p\). The \(p\)-series \(\sum_{n=1}^{\infty}1/n^p\) converges when \(p>1\) and diverges when \(0<p\leq 1\).
The degree guide predicts the absolute-value behavior, provided the rational term is eventually positive in magnitude and its leading coefficients are nonzero. It does not by itself verify that the sequence of magnitudes is decreasing, which is a separate requirement for the Alternating Series Test.
The Two Tests in the Analysis
For an alternating series whose terms have the form \((-1)^n b_n\) or \((-1)^{n+1}b_n\), the Alternating Series Test says that the series converges if \(b_n\geq 0\) eventually, \(b_n\) is nonincreasing eventually, and \(\lim_{n\to\infty}b_n=0\). A finite number of initial terms does not affect convergence, so the conditions need only hold from some index onward.
Absolute convergence is tested separately: the original series converges absolutely if the series of magnitudes \(\sum |(-1)^n b_n|=\sum b_n\) converges. For a rational \(b_n\) with \(b_n\sim c/n^p\) and \(c>0\), limit comparison with \(1/n^p\) determines this: the ratio tends to \(c\), a finite positive number, so the two positive-term series have the same convergence behavior.
Why the Degree Comparison Works
Write the leading terms of the polynomials as \(P(n)=c_Pn^{d_P}+\) lower-degree terms and \(Q(n)=c_Qn^{d_Q}+\) lower-degree terms. Divide numerator and denominator by \(n^{d_Q}\). If \(p=d_Q-d_P\), then
As \(n\) tends to infinity, the numerator on the right tends to \(c_P\), and the denominator tends to \(c_Q\). Since \(c_Q\neq 0\), the quotient tends to \(c_P/c_Q\). Thus, when the rational terms are eventually positive and \(c_P/c_Q>0\), limit comparison gives the same convergence behavior as \(\sum 1/n^p\).
This argument explains why the degree difference predicts the absolute-value series. It does not establish the alternating-series conditions: those depend on the actual sequence, including whether its magnitudes decrease eventually.
Worked Example: Conditional Convergence with a Harmonic Benchmark
Worked Example: Conditional Convergence with a Harmonic Benchmark
Determine whether the series
converges, and decide whether its convergence is absolute or conditional.
Set \(b_n=(2n+3)/(n^2+5n+6)\). For \(n\geq 1\), the numerator and denominator are positive, so \(b_n>0\). To check decrease, extend the formula to \(f(x)=(2x+3)/(x^2+5x+6)\). Its derivative is
The denominator is positive and the numerator is negative for \(x\geq 1\), so \(f\) is decreasing there; consequently, \(b_n\) decreases. Also, the denominator has degree \(2\) and the numerator degree \(1\), so \(b_n\to 0\). The Alternating Series Test therefore proves that the original series converges.
Now test absolute convergence. The degree difference is \(1\), and the leading-coefficient ratio is \(2\). Equivalently, compute the limit for comparison with the harmonic series:
Since this limit is finite and positive, \(\sum b_n\) has the same behavior as \(\sum 1/n\), which diverges. The original series converges, but not absolutely; it is conditionally convergent.
Check: The signed terms tend to zero because their magnitudes \(b_n\) tend to zero, as required by the term test. The absolute-value series diverges by the positive finite limit comparison with the harmonic series, confirming the conditional-convergence conclusion.
Worked Example: Absolute Convergence
Worked Example: Absolute Convergence
Classify the series
Let \(b_n=(5n^2+1)/(n^4+3n^2+1)\). This is positive for \(n\geq 1\). To verify that it decreases, define \(f(x)=(5x^2+1)/(x^4+3x^2+1)\). Differentiating gives
For \(x\geq 1\), \(x^3\geq x\), so \(-4x^3+4x\leq 0\), and \(-10x^5<0\). The denominator is positive, hence \(f'(x)<0\), and \(b_n\) decreases. The denominator degree exceeds the numerator degree by \(2\), so \(b_n\to 0\). The Alternating Series Test proves convergence of the signed series.
For absolute convergence, compare \(b_n\) with \(1/n^2\):
This is a positive finite limit, and \(\sum 1/n^2\) converges because \(2>1\). Therefore \(\sum b_n\) converges, so the original series converges absolutely.
Check: The derivative calculation verifies decrease for every \(x\geq 1\), and the limit comparison verifies convergence of the absolute-value series. Thus absolute convergence, which is stronger than convergence by alternation alone, is established.
Worked Example: Failure of the Term Test
Worked Example: Failure of the Term Test
Consider
The magnitude of the \(n\)th term is \(b_n=(n^2+1)/(n^2+2n+3)\). Its numerator and denominator have the same degree and the same leading coefficient, so
Thus the terms of the signed series do not tend to zero: their magnitudes tend to \(1\), rather than \(0\). The term test for divergence says that if the terms of a series do not tend to zero, the series diverges. Therefore this alternating series diverges, regardless of whether its magnitudes decrease.
The absolute-value series also diverges, since its terms tend to \(1\), not zero. The degree difference is \(0\), so there is no decaying \(p\)-series behavior that could yield absolute convergence.
Check: The original terms have magnitudes tending to \(1\), so they fail the necessary condition for convergence. The absolute-value terms have the same limit, confirming that the absolute-value series cannot converge either.
A Common Mistake: Alternation Does Not Mean Absolute Convergence
A frequent error is to see alternating signs and conclude that the series converges absolutely. Alternation can produce convergence even when the sum of the magnitudes diverges; those are different claims and require different checks.
The correction is to test each claim separately. The magnitudes \(1/n\) decrease to zero, so the Alternating Series Test gives convergence of the signed series. But the absolute-value series is \(\sum 1/n\), which is the divergent harmonic series. The alternating series is therefore conditionally convergent, not absolutely convergent.
For rational terms, degree comparison quickly signals which absolute-value benchmark to try. It does not replace the checks of eventual positivity, eventual decrease, and a zero limit needed to use the Alternating Series Test.
Check Your Understanding
For each question, distinguish the alternating-series conditions from the separate absolute-convergence test.
- For a rational magnitude with numerator degree \(2\) and denominator degree \(4\), which \(p\)-series is the natural limit-comparison benchmark for absolute convergence?
- If an alternating rational series has magnitudes decreasing eventually to zero, what conclusion does the Alternating Series Test give?
- For the series in Question 2, what additional series must be tested to determine absolute convergence?
- If a rational magnitude has numerator and denominator of the same degree, what does its leading-term behavior usually imply about the term test?
- Explain why a positive finite limit of \(b_n/(1/n^p)\) allows comparison of \(\sum b_n\) with a \(p\)-series.
