Why Rational Terms Need Two Checks
An alternating sign pattern can make a series converge even when the corresponding positive-term series does not. For rational terms, the expressions may look simple, but convergence still depends on checking two separate properties: whether the positive terms eventually decrease and whether they approach zero.
For example, the term \(1/(3n+2)\) gets smaller as \(n\) increases and tends to zero. The term \(n/(n^2+1)\) also tends to zero, but its decrease is less immediate from its appearance. In both cases, exact algebra can establish the required properties.
The purpose here is to use the Alternating-Series Test on rational-term series. The test concerns the positive magnitudes of the terms, not merely the fact that their signs alternate.
The Test and the Rational-Term Checks
For rational terms, write the series in the form \(\sum (-1)^{n-1}a_n\), where \(a_n\) is the positive magnitude of the \(n\)th term. Then check:
- Positivity: \(a_n\geq 0\) for the indices under consideration.
- Eventual decrease: \(a_{n+1}\leq a_n\) from some index onward.
- Zero limit: \(\lim_{n\to\infty}a_n=0\).
For a rational sequence, a useful exact check of decrease is to calculate \(a_{n+1}-a_n\). If the result is nonpositive for every integer \(n\) in a stated range, the sequence is nonincreasing there. The limit can often be found by comparing the degrees of the numerator and denominator.
Why Polynomial Degrees Determine the Limit
Let \(P\) and \(Q\) be nonzero polynomials with degrees \(r\) and \(s\), and leading coefficients \(p\) and \(q\), respectively. For large positive \(n\), the highest powers determine the behavior of \(P(n)/Q(n)\). Dividing the numerator and denominator by \(n^s\) makes this precise.
If \(r<s\), every term in the numerator tends to zero after this division, while the denominator tends to \(q\neq 0\). Thus \(P(n)/Q(n)\to 0\). If \(r=s\), the leading terms give the limit \(p/q\). If \(r>s\), the magnitude grows without bound, so the rational expression does not tend to zero. These conclusions assume the expression is defined for all sufficiently large positive integers \(n\).
This degree check addresses the zero-limit condition, but it does not by itself prove decrease. A rational sequence can tend to zero while increasing at some indices, so the two checks must remain separate.
Worked Examples
Worked Example: Terms of the Form \(1/(3n+2)\)
Determine whether \(\sum_{n=1}^{\infty}(-1)^{n-1}/(3n+2)\) converges by the Alternating-Series Test.
Set \(a_n=1/(3n+2)\). For every \(n\geq 1\), the denominator \(3n+2\) is positive, so \(a_n>0\). To check decrease, calculate the consecutive difference:
Both denominator factors are positive for \(n\geq 1\), so the difference is negative at every such index. Therefore \(a_n\) is strictly decreasing. For the limit, the denominator grows without bound:
The signs alternate, the magnitudes are positive and decreasing, and the magnitudes tend to zero. The Alternating-Series Test therefore shows that the series converges.
As a check on the difference calculation, \(a_1=1/5\) and \(a_2=1/8\), and \(1/8<1/5\), consistent with the general inequality.
Worked Example: Terms of the Form \(n/(n^2+1)\)
Determine whether \(\sum_{n=1}^{\infty}(-1)^{n-1}n/(n^2+1)\) converges by the Alternating-Series Test.
Set \(a_n=n/(n^2+1)\), which is positive for \(n\geq 1\). Calculate the difference with a common denominator:
The denominator is positive. For \(n\geq 1\), \(n^2+n-1\geq 1\), so the numerator \(-n^2-n+1\) is negative. Hence \(a_{n+1}<a_n\) for every \(n\geq 1\).
To find the limit, divide the numerator and denominator by \(n^2\):
Thus the positive magnitudes decrease to zero. The alternating series converges by the test. As a direct check of the first comparison, \(a_1=1/2\) and \(a_2=2/5\), and \(2/5<1/2\).
Worked Example: A Sequence That First Increases
Determine whether \(\sum_{n=1}^{\infty}(-1)^{n-1}(n+1)/(n^2+10)\) converges by the Alternating-Series Test.
Set \(a_n=(n+1)/(n^2+10)\). The numerator and denominator are positive for \(n\geq 1\), so \(a_n>0\). The first two terms are
In fact, \(3/14>2/11\), since \(3\cdot 11=33>28=2\cdot 14\). The sequence increases at this first step, so it is not decreasing from \(n=1\). Check whether it decreases from a later index by calculating:
For every integer \(n\geq 2\), \(n^2+3n-9\geq 4+6-9=1\). The denominator is positive, so \(a_{n+1}-a_n<0\) for every \(n\geq 2\). The sequence is therefore strictly decreasing from \(a_2\) onward.
For the limit, divide numerator and denominator by \(n^2\):
The Alternating-Series Test applies to the tail beginning at \(n=2\), because its magnitudes decrease to zero. Adding the single initial term does not change convergence: every partial sum after that term is the fixed number \(a_1\) plus a partial sum of the tail. Thus the original series converges.
A Complete Limit Check for Rational Terms
The degree rule can be used as a short, systematic justification whenever a rational term has polynomial numerator and denominator. For example, in \(n/(n^2+1)\), the numerator has degree \(1\) and the denominator has degree \(2\). Dividing by the denominator's highest power \(n^2\) gives
The numerator on the right tends to zero and the denominator tends to one, so the quotient tends to zero. In \((n+1)/(n^2+10)\), the same degree comparison gives a zero limit. In contrast, if numerator and denominator have the same degree, the limit is generally the ratio of their leading coefficients, not zero.
This reasoning establishes only a limit. To establish monotonicity, calculate a difference or use another valid argument. In the examples above, writing the difference over a positive denominator exposed a polynomial numerator whose sign could be checked for all indices in the required range.
Common Pitfall: Alternation Does Not Repair a Nonzero Term Limit
A common error is to see alternating signs and assume the series converges without checking the magnitudes. Consider \(\sum_{n=1}^{\infty}(-1)^{n-1}n/(3n+2)\). The magnitude is positive, and its numerator and denominator have the same degree, so
Consequently, the terms of the alternating series do not tend to zero: their magnitudes tend to \(1/3\). A necessary condition for any series to converge is that its terms tend to zero. This series therefore diverges, regardless of its alternating signs.
The incorrect reasoning would be: “The signs alternate, so the series converges.” The correction is to check the term limit first, then check eventual decrease if the limit is zero. If the term limit is nonzero, the series fails the necessary condition for convergence and there is no need to invoke the Alternating-Series Test.
A Reliable Workflow
For a rational-term alternating series, keep the checks distinct and in a deliberate order. This prevents a correct limit calculation from being mistaken for a complete convergence argument.
Write the term as an alternating sign times a positive rational sequence \(a_n\), on the indices where it is defined.
Use polynomial degrees or divide by the highest power of \(n\). If the limit is not zero, the series cannot converge.
Compute \(a_{n+1}-a_n\), identify the sign of its numerator, and state exactly from which integer index the sign holds.
If the magnitudes are eventually nonincreasing and tend to zero, conclude convergence. A finite initial segment does not change that conclusion.
Check Your Understanding
For each question, identify which convergence condition is being checked and justify your conclusion with an exact calculation.
- For \(a_n=1/(4n+1)\), compute \(a_{n+1}-a_n\) and determine whether the sequence decreases.
- For \(a_n=n/(n^2+4)\), use division by the highest power of \(n\) to find its limit.
- For \(a_n=(n+2)/(n^2+5)\), calculate the first two terms. Does this initial comparison alone decide whether the alternating series converges?
- For \(a_n=(2n+1)/(5n+3)\), determine the limit and state what it implies about the alternating series with these magnitudes.
- State the two conditions on positive magnitudes required to apply the Alternating-Series Test, allowing them to hold only from some index onward.
