Why Choose a Benchmark?
A complicated series can be hard to classify by looking at its terms alone. A useful strategy is to compare those terms with a familiar series whose convergence is already known. For nonnegative terms, two especially useful benchmarks are geometric series and p-series.
The key is not simply to find a familiar-looking expression. You must choose a benchmark with the right size and apply the comparison in the direction that supports the conclusion you want.
The Two Benchmarks
A geometric series has terms that are successive powers of a fixed ratio. For \(C>0\) and \(q\ge 0\), the series \(\sum_{n=1}^{\infty} Cq^n\) converges when \(q<1\) and diverges when \(q\ge 1\). When \(0\le q<1\), its sum is \(Cq/(1-q)\).
A p-series is a series of reciprocal powers. Its convergence classification is:
These facts let us turn a comparison into a conclusion. If \(\sum b_n\) converges and \(0\le a_n\le b_n\) for all sufficiently large \(n\), then \(\sum a_n\) converges. If \(\sum b_n\) diverges and \(a_n\ge b_n\ge0\) for all sufficiently large \(n\), then \(\sum a_n\) diverges. A finite number of initial terms does not change whether a series converges.
There is also a useful limit comparison: if \(a_n\) and \(b_n\) are positive for all sufficiently large \(n\), and
then \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge. This is especially useful when the terms are not easy to order directly but their relative sizes are clear in the limit.
Choosing Between Geometric and p-Series Comparisons
Start by identifying what controls the term size:
- Exponential factors: Terms involving \(c^n\), where \(c>1\), often suggest a geometric benchmark. A fixed power of \(n\) grows more slowly than an exponential, so polynomial factors do not prevent sufficiently rapid exponential decay.
- Reciprocal powers: Terms resembling \(1/n^p\) suggest a p-series. If a positive constant times \(1/n^p\) is the limiting size, the p-series classification supplies the conclusion.
- Rational expressions in \(n\): Compare the leading powers in numerator and denominator to identify a reciprocal-power benchmark, then verify the choice using direct or limit comparison.
These are ways to choose a candidate, not substitutes for checking the comparison conditions. The terms must be nonnegative eventually, and the comparison or limit must be established.
Worked Example: A Polynomial Factor with Exponential Decay
Worked Example: A Polynomial Factor with Exponential Decay
Determine whether \(\sum_{n=1}^{\infty}(n^2+1)/4^n\) converges.
The numerator is polynomial in \(n\), while the denominator grows exponentially. We will compare the terms with the geometric sequence \((3/4)^n\). To justify that comparison, first show that \(n^2+1\le3^n\) for every positive integer \(n\).
For \(n=1\), \(n^2+1=2\le3=3^1\). Now suppose \(n^2+1\le3^n\). The difference between \(3(n^2+1)\) and \((n+1)^2+1\) is
Thus \((n+1)^2+1\le3(n^2+1)\le3^{n+1}\), completing the induction. Therefore, term by term,
The geometric series \(\sum_{n=1}^{\infty}(3/4)^n\) converges because its ratio \(3/4\) is less than \(1\). Direct comparison therefore proves that the given series converges.
Check: At \(n=1\), the original term is \(2/4=1/2\), while the comparison term is \(3/4\), and \(1/2\le3/4\). The general inequality above verifies the comparison for every term, and the geometric benchmark meets the convergence condition.
Worked Example: A Reciprocal-Square Benchmark
Worked Example: A Reciprocal-Square Benchmark
Determine whether \(\sum_{n=1}^{\infty}(5n^2+3)/(n^4+7n+1)\) converges.
The numerator has degree \(2\) and the denominator has degree \(4\), suggesting terms of size \(1/n^2\). Set
Both sequences are positive for positive integers \(n\). Their ratio is
Divide numerator and denominator by \(n^4\):
Since the limit is finite and positive, limit comparison applies. The benchmark \(\sum 1/n^2\) converges because its p-value is \(2>1\). Thus the original series converges.
Check: At \(n=1\), \(a_1=8/9\) and \(b_1=1\), so \(a_1/b_1=8/9\). The ratio formula also gives \((5+3)/(1+7+1)=8/9\). The terms are positive, and the limiting ratio is \(5\), not \(0\) or infinity, as required for this comparison.
Worked Example: A Harmonic-Series Lower Bound
Worked Example: A Harmonic-Series Lower Bound
Determine whether \(\sum_{n=1}^{\infty}(2n+1)/(n^2+3n)\) converges.
The terms are positive for every \(n\ge1\). The expression suggests comparison with \(1/n\), the harmonic-series term. For \(n\ge2\), check the inequality directly:
Since \(n^2-2n=n(n-2)\ge0\) when \(n\ge2\), the original terms are at least \(1/n\) from that point onward. The harmonic series \(\sum 1/n\) diverges, so direct comparison proves that the given series diverges.
Check: At \(n=2\), the original term is \(5/(4+6)=1/2\), equal to \(1/n=1/2\). For every \(n\ge2\), the algebra above verifies the required lower bound. The omitted first term is finite and cannot change divergence.
Why Limit Comparison Works
The limit comparison conclusion follows from bounding the ratio near its positive finite limit. Suppose \(a_n,b_n>0\) eventually and \(a_n/b_n\to L\), where \(0<L<\infty\). By the definition of a limit, there is an index \(N\) such that for every \(n\ge N\),
This implies \(L/2<a_n/b_n<3L/2\). Multiplying by the positive \(b_n\) gives
Thus each series has terms bounded above and below by positive constant multiples of the other, apart from finitely many initial terms. The direct comparison principle then gives the same convergence behavior for both series. This explains why a ratio approaching, for example, \(5\) still makes \(1/n^2\) an effective benchmark: the terms eventually differ by only a bounded positive factor.
A Common Comparison Error
A frequent mistake is to assume that \(a_n\le b_n\) and divergence of \(\sum b_n\) imply divergence of \(\sum a_n\). The inequality goes in the wrong direction for that conclusion. For a concrete counterexample,
The larger series \(\sum 1/n\) diverges, but the smaller series \(\sum 1/n^2\) converges. A smaller nonnegative series may converge or diverge when its larger comparison series diverges. To prove divergence by direct comparison, the series being tested must instead have terms at least as large as those of a known divergent series.
A Practical Selection Process
Confirm that the terms are nonnegative, at least from some index onward, before using these comparison conclusions.
Use a geometric benchmark for exponential behavior and \(1/n^p\) for reciprocal-power behavior.
Use direct comparison when a useful inequality is clear; use limit comparison when the ratio approaches a positive finite constant.
For direct comparison, match the inequality direction to the desired conclusion. For a p-series, check whether \(p>1\).
Name the benchmark and the comparison fact that transfers its convergence or divergence to the original series.
Check Your Understanding
For each question, identify an appropriate benchmark and explain which comparison condition supports the conclusion.
- Which geometric series could serve as an upper bound for \(\sum (n^2+1)/4^n\), and why does that benchmark converge?
- For a positive series with terms whose ratio to \(1/n^3\) approaches \(7\), what convergence conclusion follows?
- What inequality direction is needed to prove divergence by comparison with the harmonic series?
- Does \(\sum 1/n^{0.8}\) converge or diverge, and which p-series rule applies?
- Why is it not valid to infer that \(\sum a_n\) diverges solely from \(0\le a_n\le b_n\) and divergence of \(\sum b_n\)?
