Why Compare Series?
A series can be difficult to sum exactly while still being possible to classify. Direct comparison gives a way to do that: if the terms of a nonnegative series are eventually no larger than the terms of a series already known to converge, then the original series converges as well.
The key work is finding and verifying the inequality. A comparison is useful only when its direction is correct, the terms satisfy the required nonnegativity condition, and the bounding series really does converge.
The Direct Comparison Theorem
For a sequence of terms \((a_n)\), the series \(\sum_{n=1}^{\infty} a_n\) converges when its sequence of partial sums converges to a finite real number. An inequality holds eventually if there is an integer \(N\) such that it holds for every \(n\ge N\).
The theorem does not require the inequality to hold for every term. A finite number of initial terms contributes only a fixed finite amount to the partial sums, so it cannot change whether a series converges. In applications, however, the terms in the tail must be nonnegative, and the comparison must be made in the direction \(a_n\le b_n\).
Why the Comparison Works
Here is a proof that makes the partial-sum mechanism explicit. Choose \(N\) so that \(0\le a_n\le b_n\) for all \(n\ge N\). For \(m\ge N\), define the partial sum of the \(a_n\) tail by \(A_m=\sum_{n=N}^{m}a_n\). Term-by-term comparison gives
The final upper bound is finite because \(\sum b_n\) converges. Also, \(A_m\) is nondecreasing: increasing \(m\) adds a nonnegative term. A nondecreasing sequence of real numbers that is bounded above converges. Thus the partial sums of the \(a_n\) tail converge. Adding the fixed finite sum of the terms before \(N\) shows that the full partial sums of \(\sum a_n\) converge too. This proves the theorem.
Choosing a Comparison Series
In many problems, a suitable comparison series is a familiar one. For example, a constant multiple of a p-series \(\sum 1/n^p\) converges when \(p>1\), and a geometric series with ratio whose absolute value is less than \(1\) converges. A telescoping series can also provide a useful bound when its partial sums have a finite limit. The comparison need not be the closest possible bound; it only needs to be valid and have a known convergent sum.
Identify where the terms are nonnegative and where the proposed inequality will hold.
Find a sequence \(b_n\) with \(a_n\le b_n\) for every sufficiently large \(n\).
Give a reason that \(\sum b_n\) converges, such as a p-series or geometric-series result.
Apply the theorem to conclude that \(\sum a_n\) converges.
Worked Examples
Worked Example: A Rational Term Compared with a p-Series
Show that \(\sum_{n=1}^{\infty}\frac{3n+1}{n^3+2n}\) converges.
For \(n\ge1\), \(3n+1\le4n\), since \(4n-(3n+1)=n-1\ge0\). Also \(n^3+2n\ge n^3>0\). Therefore, using positive denominators,
The comparison series \(\sum_{n=1}^{\infty}4/n^2=4\sum_{n=1}^{\infty}1/n^2\) converges because it is a constant multiple of a p-series with exponent \(2>1\). The direct comparison theorem therefore gives convergence of the original series.
Check: The original terms are positive, the displayed inequality holds for every \(n\ge1\), and the upper-bound series converges. All the theorem’s conditions are met.
Worked Example: An Inequality That Holds Only Eventually
Define \(a_1=5\), and for \(n\ge2\) define \(a_n=\frac{1}{n(n-1)}\). Show that \(\sum_{n=1}^{\infty}a_n\) converges.
The first term is a finite exception. For \(n\ge2\), \(n-1\ge n/2\), because \(n-2\ge0\). Multiplying by \(n>0\) gives \(n(n-1)\ge n^2/2>0\), so
The series \(\sum_{n=2}^{\infty}2/n^2\) converges by the p-series test. The comparison is valid for every \(n\ge2\), so the direct comparison theorem proves convergence of the tail \(\sum_{n=2}^{\infty}a_n\). Including \(a_1=5\) adds just one finite term and does not change convergence.
Check: At \(n=2\), \(a_2=1/2\) and \(2/n^2=1/2\), so the boundary case satisfies the inequality with equality. The exceptional first term is finite, and the comparison series converges.
Worked Example: Comparison with a Geometric Series
Show that \(\sum_{n=1}^{\infty}\frac{n+1}{4^n}\) converges.
First, \(n+1\le2^n\) for all positive integers \(n\). This holds with equality at \(n=1\). If it holds for \(n\), then \(n+2\le2(n+1)\le2\cdot2^n=2^{n+1}\), so induction establishes the inequality for all \(n\ge1\). Since \(4^n>0\), it follows that
The comparison series \(\sum_{n=1}^{\infty}(1/2)^n\) is geometric with ratio \(1/2\), whose absolute value is less than \(1\). Its sum is \((1/2)/(1-1/2)=1\), so it converges. Direct comparison proves that the given series converges.
Check: The numerator bound holds for every positive integer, and dividing it by the positive quantity \(4^n\) preserves the inequality. The resulting upper bound is a convergent geometric series.
Worked Example: Comparison with a Telescoping Series
Show that \(\sum_{n=1}^{\infty}\frac{1}{n(n+2)}\) converges.
For every \(n\ge1\), \(n+2\ge n+1>0\), so \(n(n+2)\ge n(n+1)>0\). Taking reciprocals reverses the denominator inequality:
The comparison series telescopes because \(1/[n(n+1)]=1/n-1/(n+1)\). Its \(m\)-th partial sum is
As \(m\) tends to infinity, \(1/(m+1)\) tends to zero, so these partial sums tend to \(1\). The comparison series converges, and the direct comparison theorem proves convergence of the original series.
Check: The inequality holds for every \(n\ge1\), and the upper-bound partial sums tend to the finite value \(1\). Thus the original series satisfies the theorem’s conditions.
Common Mistake: Reversing the Comparison
For convergence, the series being tested must be bounded above by a convergent series. A convergent lower bound does not establish convergence. For a concrete example, \(1/n\ge1/n^2\) for every \(n\ge1\), and \(\sum 1/n^2\) converges. It would be incorrect to infer that \(\sum 1/n\) converges from this inequality; the inequality points in the wrong direction. In fact, the harmonic series \(\sum 1/n\) diverges.
What the Conclusion Does—and Does Not—Say
The theorem establishes convergence from a valid upper bound. It does not provide the exact sum of \(\sum a_n\), and it does not require the comparison to be sharp. It also does not say that \(\sum a_n\) and \(\sum b_n\) have equal sums. Their terms may differ substantially while the upper bound still suffices to prove convergence.
When writing a comparison argument, include the inequality, the range of indices on which it holds, the reason the comparison series converges, and the conclusion. This makes clear both how the terms are bounded and why the theorem applies.
Check Your Understanding
For each question, identify the comparison inequality and verify that the theorem’s conditions are satisfied.
- For \(n\ge1\), show that \(\frac{1}{n^2+3n}\le\frac{1}{n^2}\). What does this imply about the series with terms \(1/(n^2+3n)\)?
- Suppose \(0\le a_n\le 5/n^3\) for all \(n\ge4\). What fact about the comparison series lets you conclude that \(\sum a_n\) converges?
- Why can a finite number of terms before the comparison becomes valid be ignored when deciding convergence?
- What is wrong with trying to prove convergence of \(\sum a_n\) from \(1/n^2\le a_n\) and the convergence of \(\sum 1/n^2\)?
- In the geometric-series example, what inequality bounds \(n+1\), and why does dividing by \(4^n\) preserve its direction?
