Why Several Bounds May Be Available
A positive term can often be bounded in more than one simple way. One bound may be too weak to decide whether its series converges, while another bound for the same terms may settle the question immediately. The key is not to find the closest possible estimate. It is to find a bound with the right direction and a comparison series whose behavior is already known.
For convergence, we seek an upper bound by the terms of a convergent series. For divergence, we seek a lower bound by the terms of a divergent series. The same inequality cannot generally serve both purposes: the direction of comparison determines what conclusion follows.
The Direction of the Bound Determines the Conclusion
The direct comparison rules use the fact that partial sums of nonnegative terms cannot decrease. If a positive-term series is bounded above term by term by a convergent series, its partial sums stay bounded. If it is bounded below term by term by a divergent series, its partial sums grow without bound.
- If \(a_n\le b_n\) eventually and \(\sum b_n\) converges, then \(\sum a_n\) converges.
- If \(a_n\ge b_n\) eventually and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.
Here is the partial-sum reason behind the two rules. If \(a_n\le b_n\) for \(n\ge N_0\), then, for \(N\ge N_0\), the partial sum of the \(a_n\) terms through \(N\) is at most its fixed initial portion plus the corresponding partial sum of the \(b_n\) terms. The latter is bounded because \(\sum b_n\) converges. The partial sums of \(a_n\) are nondecreasing and bounded, so they converge. Conversely, if \(a_n\ge b_n\) from \(N_0\) onward and the partial sums of \(b_n\) grow without bound, the partial sums of \(a_n\) also grow without bound. This explains why the upper bound proves convergence and the lower bound proves divergence.
In the examples, we will use the \(p\)-series test: \(\sum 1/n^p\) converges when \(p>1\) and diverges when \(p\le1\). In particular, \(\sum 1/n^2\) converges and the harmonic series \(\sum 1/n\) diverges.
A Practical Way to Choose a Comparison
When several bounds seem possible, begin by deciding which conclusion the evidence should support. Then choose a comparison series that can actually establish that conclusion. To prove convergence, look for an upper bound with a power \(p>1\), or another known convergent comparison. To prove divergence, look for a lower bound with a power \(p\le1\), or another known divergent comparison.
Ask whether the expression suggests a useful upper bound for convergence or a useful lower bound for divergence.
For each candidate, check that the direction is correct and that the inequality holds for every index in the stated range.
Use a known convergence or divergence result. If that result gives no conclusion in the required direction, try a different bound.
Explain which direct comparison rule applies and why its hypotheses hold.
Worked Example: Choose the Stronger Upper Bound
Worked Example: Choose the Stronger Upper Bound
Determine whether the series \(\sum_{n=1}^{\infty}\frac{3n+2}{n^3+4n^2+1}\) converges.
All numerator and denominator terms are positive for \(n\ge1\). First compare the numerator with \(5n\). Since \(n\ge1\), we have \(2\le2n\), so \(3n+2\le5n\). Also, \(n^3+4n^2+1\ge n^3\). Therefore, dividing positive quantities in the indicated directions gives
The comparison series \(\sum 5/n^2\) converges, because it is \(5\) times a convergent \(p\)-series with \(p=2\). By direct comparison, the given series converges.
A weaker upper bound is also available. Since \(n^3+4n^2+1\ge4n^2\) and \(3n+2\le5n\), we get
This inequality is correct, but \(\sum 5/(4n)\) diverges. An upper bound by a divergent series does not prove convergence or divergence of the original series. The \(5/n^2\) bound is the useful choice. As a numerical check at \(n=1\), the original term is \(5/6\), while \(5/n^2=5\) and \(5/(4n)=5/4\); the stated upper bounds both hold.
Worked Example: Choose a Lower Bound That Diverges
Worked Example: Choose a Lower Bound That Diverges
Determine whether the series \(\sum_{n=1}^{\infty}\frac{2n+1}{n^2+4n+3}\) diverges.
For \(n\ge1\), \(2n+1\ge2n\). Also \(n\le n^2\) and \(1\le3n^2\), so
All quantities are positive. The numerator is at least \(2n\), and the denominator is at most \(8n^2\), so
The comparison series \(\sum 1/(4n)\) diverges because it is one quarter of the harmonic series. The original series therefore diverges by direct comparison.
There is also a true, but less useful, lower bound. Since \(2n+1\ge1\) and the denominator is at most \(8n^2\), we have
The series \(\sum 1/(8n^2)\) converges. A lower bound by a convergent series does not establish convergence of the original series; the original terms might be larger. Thus the harmonic lower bound, not merely any valid lower bound, is the one that proves divergence. At \(n=1\), the original term is \(3/8\), and the two lower bounds are \(1/4\) and \(1/8\), respectively, confirming both inequalities at this index.
Worked Example: Use Two Bounds to Reveal Divergence
Worked Example: Use Two Bounds to Reveal Divergence
Determine whether the series \(\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+1}+n}\) converges.
For \(n\ge1\), \(n^2+1\ge n^2\), so \(\sqrt{n^2+1}\ge n\). Thus the denominator is at least \(2n\), giving an upper bound. In the other direction, \(n^2+1\le2n^2\), so \(\sqrt{n^2+1}\le\sqrt{2}\,n<2n\). The denominator is therefore less than \(3n\), giving a lower bound. Together, these inequalities yield
The upper bound is by a constant multiple of the harmonic terms and does not prove convergence. The lower bound is by a positive constant multiple of those same terms, and that comparison series diverges. Hence the given series diverges by direct comparison.
For \(n=1\), the term is \(1/(\sqrt{2}+1)\). Since \(1<\sqrt{2}<2\), its denominator is between \(2\) and \(3\), so the term is between \(1/3\) and \(1/2\), as the displayed inequalities assert. The bounds need not be sharp: their purpose is to provide a valid lower comparison with a known divergent series.
Common Pitfall: A Bound Is Not a Conclusion by Itself
A frequent error is to find a valid inequality and then draw a conclusion without checking whether the comparison series has the needed behavior. For instance, suppose \(0\le a_n\le 5/n\). It is incorrect to conclude that \(\sum a_n\) converges merely because every \(a_n\) is at most \(5/n\). The comparison series \(\sum 5/n\) diverges, so the upper-bound rule does not apply.
Similarly, if \(a_n\ge1/(8n^2)\), it is incorrect to conclude that \(\sum a_n\) converges because \(\sum1/(8n^2)\) converges. The inequality is a lower bound, and a sequence can be larger than terms of a convergent series while its own series diverges. Before using a comparison, name the direction, name the comparison series, and verify that the relevant direct comparison rule gives the desired conclusion.
Why Comparing Several Bounds Matters
Expressions can contain sums, powers, or radicals that allow several elementary estimates. Different estimates may produce different powers of \(n\), and the \(p\)-series threshold makes that difference decisive. A bound by \(C/n\) may be too weak to prove convergence, while a bound by \(C/n^2\) can settle it. For divergence, the useful direction reverses: a lower bound by \(c/n\) is decisive, but a lower bound by \(c/n^2\) is not.
It is also important to keep the quantifiers in view. A comparison need only hold eventually, meaning for every \(n\) beyond some fixed index. If an inequality holds only for \(n\ge N_0\), the finitely many earlier terms do not change the behavior of the infinite sum. State the range where the bound is valid, and make sure its terms are nonnegative there.
Check Your Understanding
For each question, focus on both the direction of the inequality and the behavior of the comparison series.
- If \(0\le a_n\le b_n\) eventually and \(\sum b_n\) converges, what can be concluded about \(\sum a_n\)?
- Why does an upper bound by \(1/n\) fail to prove convergence of a positive-term series?
- If \(a_n\ge1/(5n)\) eventually, what conclusion follows, and which comparison fact is used?
- Can a lower bound by \(1/n^2\) establish convergence? Explain why or why not.
- Does a direct comparison have to hold for every index starting at \(1\), or is it enough for all sufficiently large indices?
