Why Build Termwise Inequalities?
A series may be difficult to recognize at first glance. A useful step is to replace its terms with simpler expressions that are easier to compare. The goal is not to guess a bound and hope it works: it is to construct an inequality whose direction, conditions, and usefulness are all clear.
For positive-term series, an upper bound can help establish convergence, while a lower bound can help establish divergence. This tutorial focuses on building those bounds by simplifying fractions, using positive denominator terms carefully, and comparing powers of \(n\).
The word “eventual” matters. A comparison need not hold for the first few terms. Changing or removing finitely many terms does not change whether a series converges or diverges, so a bound valid from some index onward can still be useful.
Choose the Direction Before Doing Algebra
Suppose \(a_n\) is positive. To prove \(a_n\leq b_n\), seek an expression that is at least as large as \(a_n\). To prove \(a_n\geq b_n\), seek one that is at most as large as \(a_n\). This distinction is particularly important when the term is a fraction.
If \(P\) and \(Q\) are positive, then making a denominator larger makes its reciprocal smaller:
More generally, for positive numerators and denominators, cross-multiplication preserves the inequality after multiplying by the positive denominators. If the denominators are not known to be positive, cross-multiplication may reverse an inequality or fail to be valid; establish positivity first.
Using Positivity and Comparing Powers
If a denominator is a sum of positive terms, discarding some of those terms makes it smaller. For example, \(n^3+2n^2+7\geq n^3\) for \(n\geq1\). Consequently, a positive fraction with this denominator is no larger than the fraction obtained by replacing the denominator with \(n^3\), provided the numerator is unchanged.
Likewise, for \(n\geq1\), higher powers are at least as large as lower powers: \(n^3\geq n^2\geq n\geq1\). These comparisons let us simplify numerators and denominators. But they must be used in the correct direction: increasing a positive numerator increases a fraction, whereas increasing a positive denominator decreases it.
Check that the terms and denominators are positive on the index range being used.
Decide whether the goal is an upper or lower bound before simplifying.
Use inequalities for the numerator and denominator, keeping track of how each change affects the fraction.
Verify the inequality algebraically or by cross-multiplication, and state the indices for which it holds.
Worked Examples
Worked Example: An Upper Bound from a Polynomial Denominator
For \(n\geq1\), find a simple upper bound for
First, the denominator is positive. Since \(n\geq1\), we have \(n-1\geq0\), so
The positive denominator also satisfies
Increasing the numerator and decreasing the denominator both produce an upper bound. Therefore,
Both component inequalities hold for every \(n\geq1\), so the resulting bound does too. As an application, the series with terms \(a_n\) converges by direct comparison with \(4\sum_{n=1}^{\infty}1/n^2\), which converges by the \(p\)-series criterion with \(p=2\).
Check: At \(n=1\), \(a_1=4/10=2/5\), while \(4/n^2=4\), so the proposed inequality holds at the first index as well. The proof above establishes it for all \(n\geq1\), not just this check.
Worked Example: A Lower Bound from a Quadratic Denominator
Show that
for all integers \(n\geq3\). The numerator, denominator, and \(n\) are positive, so cross-multiplication is valid. The proposed inequality is equivalent to
Subtracting the right-hand side from the left gives
At \(n=3\), this difference is \(2(9)-9-7=2>0\). Moreover, if \(f(n)=2n^2-3n-7\), then
for \(n\geq3\). Thus \(f(n)\geq f(3)>0\) for every integer \(n\geq3\), proving the claimed lower bound. The series \(\sum b_n\) therefore diverges: its tail is at least the corresponding tail of the harmonic series \(\sum 1/n\), which diverges.
Check: At \(n=3\), \(b_3=11/31\), and \(1/3\leq11/31\) because \(31\leq33\). The algebraic proof verifies the bound for every \(n\geq3\).
Worked Example: Two Bounds by Comparing Powers
Find upper and lower bounds for
for every integer \(n\geq1\). For the upper bound, \(n^2\geq1\), so \(3\leq3n^2\). Hence \(2n^2+3\leq5n^2\). Also \(5n^3+n\geq5n^3>0\). It follows that
For the lower bound, \(n\leq n^3\) for \(n\geq1\), so \(5n^3+n\leq6n^3\). The numerator satisfies \(2n^2+3\geq2n^2\). Making the numerator smaller and the denominator larger gives a lower bound:
Thus both estimates hold for every \(n\geq1\):
Check: At \(n=1\), \(c_1=5/6\), and \(1/3\leq5/6\leq1\). The general proof comes from the stated numerator and denominator inequalities, which hold throughout the required range.
The lower estimate also shows that the series with terms \(c_n\) diverges, because \(c_n\geq(1/3)(1/n)\) and a positive constant multiple of the harmonic series diverges.
Why the Comparison Argument Works
The algebra that produces a bound is only part of the reasoning. The bound must point in a direction that supports a valid conclusion about the series. For nonnegative terms, the partial sums preserve termwise order: if \(0\leq a_n\leq b_n\) for every \(n\geq N\), then for \(m\geq N\),
If the partial sums of the upper comparison series remain bounded, the partial sums of the series with terms \(a_n\) remain bounded as well, and the latter series converges. In the other direction, if \(a_n\geq b_n\geq0\) eventually and the partial sums of \(\sum b_n\) grow without bound, then the partial sums of \(\sum a_n\) also grow without bound. These are the direct comparison rules for positive-term series.
The key condition is that the inequalities hold for all terms from some index onward, not merely for selected values of \(n\). A handful of numerical checks can catch arithmetic errors, but it cannot establish an eventual bound.
A Common Mistake: Reversing the Effect of a Denominator
Consider \(d_n=(n+1)/(n^2+n+1)\), for \(n\geq1\). A mistaken argument might say that \(n^2+n+1\geq n^2\), so
This conclusion has the inequality backwards. The numerator is positive, and replacing the denominator by the smaller quantity \(n^2\) makes the fraction larger, not smaller. For example, at \(n=1\), the original term is \(2/3\), while the proposed lower bound is \(2\), so the claim fails.
The correct conclusion from \(n^2+n+1\geq n^2>0\) is
For a lower bound, one instead needs an upper estimate for the denominator, along with a suitable lower estimate for the numerator. Every time a denominator is simplified, pause to ask whether it has been made larger or smaller and what that does to the fraction.
Building a Bound That Is Useful
Not every correct inequality helps answer the question. A bound is useful when it leads to a comparison series whose behavior is already known. For convergence, seek an eventual upper bound by a convergent positive-term series. For divergence, seek an eventual lower bound by a divergent positive-term series.
Keep the scope of each estimate explicit. For example, \(n^2\geq n\) requires \(n\geq1\), while an inequality derived from a polynomial difference may only hold beyond a larger threshold. State that threshold, and do not silently apply the estimate before it begins to hold.
Check Your Understanding
For each question, identify the bound direction and justify the range of indices on which it holds.
- For \(n\geq1\), why does \(\frac{2n+1}{n^2+4}\leq\frac{3}{n}\) follow from simple numerator and denominator estimates?
- When a positive denominator is replaced by a smaller positive expression, does the resulting fraction increase or decrease?
- Verify by cross-multiplication whether \(\frac{4n+1}{n^2+3n+2}\geq\frac{1}{n}\) holds for every \(n\geq1\).
- Why can a termwise lower bound by a positive constant multiple of \(1/n\) establish divergence of a positive-term series?
- Give one reason a bound that holds only for \(n\geq N\) can still be used in a series comparison.
