When a Lower Bound Proves Divergence
A useful way to prove that a series diverges is to show that its terms are at least as large as the terms of a series already known to diverge. The key is the direction of the inequality: a series of nonnegative terms cannot have smaller partial sums than a divergent series if each of its terms is at least as large.
The comparison does not need to hold from the very first term. A finite number of initial terms cannot determine whether a series converges or diverges. It is enough for the lower bound to hold from some index onward, provided the terms being compared are nonnegative there.
The Direct Comparison Test for Divergence
The benchmark series must have nonnegative terms and be known to diverge. Common benchmarks include the harmonic series \(\sum_{n=1}^{\infty}1/n\) and the p-series \(\sum_{n=1}^{\infty}1/n^p\) for \(0<p\leq 1\). The harmonic series is the p-series with \(p=1\).
The condition that the comparison holds for all \(n\geq N\) is sometimes called an eventual inequality. The terms before \(N\) may be handled separately; there are only finitely many of them.
Why the Test Works
For every \(m\geq N\), the comparison of individual terms gives a comparison of the corresponding finite sums. This is the central reason the test works.
Because the benchmark series diverges and its terms are nonnegative, its partial sums from \(N\) onward grow without bound. Therefore the partial sums of the \(a_n\) terms from \(N\) onward also grow without bound. Adding the fixed initial sum \(\sum_{n=1}^{N-1}a_n\) cannot change that conclusion.
Make sure the terms being compared are nonnegative from some index onward.
A harmonic or divergent p-series is often convenient.
Prove \(a_n\geq b_n\) for every \(n\) from a specific index onward.
The partial sums of the original series eventually dominate those of the divergent benchmark.
Worked Example: A Rational Term Bounded Below by a Harmonic Term
Worked Example: A Rational Term Bounded Below by a Harmonic Term
Determine whether the series \(\sum_{n=1}^{\infty}\frac{2n+1}{n^2+3n+1}\) converges or diverges.
For \(n\geq 1\), the denominator \(n^2+3n+1\) is positive, as is \(2n+1\). We compare the term with \(1/(2n)\). Since both denominators are positive, the desired inequality is equivalent to cross-multiplying:
Expanding and subtracting the right side gives
For \(n\geq1\), \(3n^2-n-1\geq 3n-n-1=2n-1\geq1>0\), where \(n^2\geq n\). Thus the termwise lower bound holds for every positive integer \(n\).
The benchmark series is divergent because
is one-half of the divergent harmonic series. Therefore, by direct comparison for divergence, the original series diverges to \(+\infty\).
Check: The cross-multiplied difference is positive for every \(n\geq1\), and the lower bound is a positive constant multiple of the harmonic term. Both conditions needed for the comparison are satisfied.
Worked Example: A p-Series Lower Bound
Worked Example: A p-Series Lower Bound
Determine whether \(\sum_{n=1}^{\infty}\frac{1}{\sqrt n+1}\) converges or diverges.
For \(n\geq1\), \(\sqrt n\geq1\), so \(1\leq\sqrt n\). Adding \(\sqrt n\) to both sides yields
Both sides are positive, so taking reciprocals reverses the inequality:
The comparison series is a positive constant multiple of the p-series with \(p=\frac12\). Since \(0<\frac12\leq1\), this p-series diverges. Consequently, the given series diverges to \(+\infty\).
Check: The denominator inequality holds for every \(n\geq1\), and reciprocation is valid because both denominators are positive. The benchmark has exponent \(p=\frac12\), within the divergent p-series range.
Worked Example: A Bound That Starts Later
Worked Example: A Bound That Starts Later
Determine whether \(\sum_{n=1}^{\infty}\frac{n+2}{n^2+5n+4}\) converges or diverges.
The numerator and denominator are positive for every positive integer \(n\). We seek a lower bound of \(1/(2n)\). For positive denominators, the comparison is equivalent to
The difference between the two sides is
At \(n=3\), this difference is \(3^2-3-4=2>0\). For each integer \(n\geq3\), its increase from \(n\) to \(n+1\) is
It follows that \(n^2-n-4\geq2>0\) for every integer \(n\geq3\). Thus the term comparison holds from \(n=3\) onward. The first two terms form only a finite sum and do not affect convergence. The tail is bounded below by \(\sum_{n=3}^{\infty}1/(2n)\), which diverges because it is a constant multiple of a tail of the harmonic series. Therefore, the original series diverges to \(+\infty\).
Check: The comparison is established for every integer \(n\geq3\), not merely for a few tested values. Omitting the first two terms leaves a divergent harmonic tail, so the eventual bound is sufficient.
Common Mistakes
The most frequent error is reversing the inequality needed for a divergence comparison. A lower bound by a divergent series can force divergence; an upper bound by a divergent series cannot.
Common Mistake: Using an Upper Bound
A mistaken argument might say: “Since \(1/n^2\leq1/n\), and \(\sum 1/n\) diverges, \(\sum 1/n^2\) diverges.” The inequality points in the wrong direction for that conclusion. In fact, \(\sum 1/n^2\) converges, as follows from the p-series criterion with \(p=2\).
The direct comparison for divergence would require a divergent series below \(1/n^2\), not above it. Always write the inequality in the form \(a_n\geq b_n\), with \(\sum b_n\) known to diverge, before drawing the conclusion.
A second important check is nonnegativity. The partial-sum proof relies on adding inequalities without reversing their direction. If the terms of the series being studied are negative or change sign in the tail, the direct comparison theorem stated here does not apply as written.
How to Use the Test Reliably
Comparison choices usually come from simplifying the expression enough to see a familiar benchmark. For a rational expression, cross-multiplication may give a direct way to prove an inequality. For expressions involving radicals, bounding a denominator from above can produce a lower bound for its reciprocal. In each case, positivity must be checked before multiplying or taking reciprocals.
It is also useful to separate the algebra from the conclusion. First prove the inequality for a stated range of \(n\); then identify the divergent benchmark; only then apply the theorem. If the inequality begins at \(n=N\), explicitly note that the earlier terms are finite in number. This avoids the mistaken belief that a comparison must start at \(n=1\).
Check Your Understanding
For each question, identify the inequality direction and the benchmark needed to justify a divergence conclusion.
- What conditions on \(a_n\) and \(b_n\) allow the direct comparison test to prove that \(\sum a_n\) diverges?
- Why does \(a_n\geq b_n\) for every \(n\geq N\) suffice even if the inequality fails for some earlier terms?
- For \(n\geq1\), show that \(\frac{1}{n+\sqrt n}\geq\frac{1}{2n}\), and identify the divergent benchmark.
- Why does the inequality \(1/n^2\leq1/n\) not prove that \(\sum 1/n^2\) diverges?
- What role does nonnegativity play in the partial-sum argument for direct comparison?
