Introduction
A convergence test is useful only when it matches the structure of the series and its hypotheses are satisfied. In a mixed set of problems, the main task is often not complicated algebra; it is recognizing which feature of the terms is most informative.
Rational terms invite comparison with a power series, factorials often simplify in a ratio, and alternating signs call for checking the conditions of the Alternating Series Test. The examples below show how to make those choices and how to state a conclusion supported by the calculation.
A Short Reference for Choosing a Test
For a series \(\sum_{n=1}^{\infty} a_n\), convergence means that its sequence of partial sums approaches a finite limit. A necessary condition is that \(a_n\to 0\). If the terms do not approach zero, the series diverges; if they do approach zero, that fact alone does not establish convergence.
These tests are especially useful for the patterns in this workshop:
| Structure | Useful test | Decisive information |
|---|---|---|
| Positive rational terms | Limit comparison | Compare with \(1/n^p\) using the leading powers. |
| Factorials and exponentials | Ratio Test | Calculate the limit of consecutive absolute-term ratios. |
| Alternating terms | Alternating Series Test | Check that magnitudes decrease to zero. |
For positive terms \(a_n\) and \(b_n\), the Limit Comparison Test applies when the ratio \(a_n/b_n\) has a finite positive limit. The two series then either both converge or both diverge. For a series with terms of both signs, absolute convergence means that \(\sum |a_n|\) converges; conditional convergence means that \(\sum a_n\) converges but \(\sum |a_n|\) diverges.
Why Limit Comparison Works
Here is the reason. Since \(a_n/b_n\to L\), the ratio is eventually close enough to \(L\) to lie between \(L/2\) and \(3L/2\). Thus, for all sufficiently large \(n\),
These inequalities compare the tails of the two positive series. If \(\sum b_n\) converges, then the upper bound shows that \(\sum a_n\) converges by direct comparison. If \(\sum a_n\) converges, the lower bound gives \(b_n\leq (2/L)a_n\), so \(\sum b_n\) converges as well. Therefore either both converge or both diverge. Adding or removing finitely many terms does not change convergence.
Worked Examples: Rational Series
Worked Example: A Convergent Positive Rational Series
Classify \(\displaystyle\sum_{n=1}^{\infty}\frac{5n+2}{n^3+2n}\).
The terms are positive. The numerator has degree \(1\), while the denominator has degree \(3\), suggesting comparison with \(1/n^2\). Set
Calculate the ratio:
The limit is finite and positive, so the Limit Comparison Test applies. The benchmark \(\sum 1/n^2\) is a convergent \(p\)-series, since \(p=2>1\). Therefore the given series converges. As a check on the comparison, the ratio approaches the positive constant \(5\), rather than zero or infinity; the two terms have the same order for large \(n\).
Worked Example: A Divergent Positive Rational Series
Classify \(\displaystyle\sum_{n=1}^{\infty}\frac{3n^2+1}{n^3+4n}\).
Again the terms are positive. The degrees suggest a term of order \(1/n\), so compare with the harmonic series:
Then
The positive finite limit permits limit comparison. Since \(\sum 1/n\) is the divergent harmonic series, the given series diverges. The conclusion is about the sum of the terms: although an individual term becomes small, the terms behave like a positive constant times \(1/n\), whose sum does not remain finite.
Worked Examples: Factorials and Alternating Terms
Worked Example: A Convergent Factorial Series
Classify \(\displaystyle\sum_{n=1}^{\infty}\frac{4^n}{n!}\).
For positive terms involving both powers and factorials, the Ratio Test is efficient. Let \(a_n=4^n/n!\). Then
Therefore
The Ratio Test says that a series of positive terms converges when this limit is less than \(1\). Hence the series converges. More concretely, for \(n\geq 7\), \(4/(n+1)\leq 1/2\); each term after that point is at most half the preceding term, so the tail is bounded by a convergent geometric series. This agrees with the test’s conclusion.
Worked Example: A Divergent Factorial Series
Classify \(\displaystyle\sum_{n=1}^{\infty}\frac{n!}{5^n}\).
Set \(a_n=n!/5^n\), which is positive. Its consecutive-term ratio is
This ratio tends to infinity, so it is eventually greater than \(1\). In particular, for \(n\geq 5\), it is at least \(6/5\). Thus the terms from that point onward increase by a factor of at least \(6/5\); they cannot approach zero. Since a convergent series must have terms tending to zero, the series diverges. The Ratio Test also gives divergence because its limiting ratio is greater than \(1\).
Worked Example: Conditional Convergence in an Alternating Rational Series
Classify \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2n+1}{n(n+1)}\).
Write the positive term magnitude as
Both summands decrease with \(n\), so \(b_n\) decreases. Explicitly,
Also, \(b_n=1/n+1/(n+1)\to 0\). The Alternating Series Test therefore establishes convergence of the original series.
To determine whether convergence is absolute, examine the series of magnitudes. For every \(n\geq1\),
because \(2n+1\geq n+1\). The harmonic series diverges, so direct comparison shows that the absolute-value series diverges. The original series is therefore conditionally convergent, not absolutely convergent. The test conditions and the comparison establish both parts of the classification.
A Pitfall: A Ratio Limit of One Is Not a Verdict
The Ratio Test proves convergence if the limit of consecutive absolute-term ratios is less than \(1\), and divergence if it is greater than \(1\) (or is infinite). If the limit equals \(1\), the test is inconclusive: it does not classify the series.
For example, for the harmonic series, \(a_n=1/n\), and the ratio is
A mistaken conclusion would be: “The ratio limit is \(1\), so the series converges.” That conclusion does not follow from the Ratio Test. In fact, the harmonic series diverges. The correction is to record that the test gives no conclusion here and use a test that does decide the case, such as recognizing the harmonic series itself.
A Reliable Classification Routine
Look for positivity, alternating signs, factorials, exponentials, or rational expressions.
For rational terms, use leading powers to select a \(p\)-series; for factorials, simplify consecutive ratios; for alternating terms, check the magnitude sequence.
Confirm positivity where comparison is used, or decrease and limit zero for the Alternating Series Test.
For alternating series, separately check absolute convergence before calling the result conditional.
Why the Choice of Test Matters
Series can look complicated while having terms with a familiar dominant pattern. A rational expression may reduce to a multiple of \(1/n^p\) for comparison purposes, and a factorial may simplify dramatically when divided by its neighboring term. Choosing a test that exposes that pattern avoids unnecessary algebra.
The reverse is equally important: a calculation is not a conclusion unless its conditions apply. Limit comparison needs a positive finite ratio limit; the alternating test requires decreasing magnitudes tending to zero; and the Ratio Test gives no verdict when its limit is \(1\). A good solution makes the choice, the condition, and the decisive calculation visible.
Check Your Understanding
For each series, select a useful test and identify the calculation or condition that decides the classification.
- Classify \(\displaystyle\sum_{n=1}^{\infty}\frac{2n+7}{n^3+n}\) by comparison with a \(p\)-series.
- For \(\displaystyle\sum_{n=1}^{\infty}\frac{3^n}{n!}\), what consecutive-term ratio should be calculated, and what does its limit imply?
- For \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}/\sqrt{n}\), which conditions must be checked to apply the Alternating Series Test?
- Why does a Ratio Test limit of \(1\) fail to establish convergence or divergence?
- What additional series must be examined after an alternating series converges, if the goal is to decide whether convergence is absolute or conditional?
