Why Compare Series?
A complicated series can be difficult to evaluate directly. Sometimes, however, its terms are visibly no larger than the terms of a simpler series whose convergence is already known. If both sets of terms are nonnegative, this upper bound can control every partial sum of the complicated series. That is the idea behind the Direct Comparison Test for Convergence.
The test is useful when a series resembles a geometric series or a p-series, but its exact terms are less familiar. The essential work is not merely spotting a resemblance: you must establish the correct inequality and make sure the comparison series converges.
In this tutorial, the series being tested and the comparison series have nonnegative terms. That condition is important: it ensures the partial sums increase as more terms are added.
The Direct Comparison Test
It is enough for the inequalities to hold from some index onward. A finite number of initial terms cannot change whether a series converges: adding or removing finitely many finite terms changes partial sums by only a fixed amount.
The comparison series is commonly a geometric series or a p-series. A geometric series with ratio \(r\) converges when \(|r|<1\); a p-series \(\sum_{n=1}^{\infty}1/n^p\) converges when \(p>1\). These provide useful benchmarks for comparison.
Why the Test Works
For nonnegative terms, every partial sum is at least as large as the preceding one. If the terms \(a_n\) are also bounded above by the terms \(b_n\), then each partial sum of the first series is bounded above by a partial sum of the second.
Here is the full argument when the comparison holds for every \(n\). Let \(S_N=\sum_{n=1}^{N}a_n\). Since \(a_n\ge0\), we have \(S_{N+1}=S_N+a_{N+1}\ge S_N\), so the sequence \((S_N)\) is nondecreasing. Since \(a_n\le b_n\), each partial sum satisfies
The final quantity is finite because \(\sum b_n\) converges. Thus \((S_N)\) is nondecreasing and bounded above, so it has a finite limit. By the definition of series convergence, \(\sum a_n\) converges.
If the inequality starts only at \(n=N_0\), apply the same argument to the tail \(\sum_{n=N_0}^{\infty}a_n\). The terms before \(N_0\) contribute only a finite sum, so adding them back still gives a convergent series.
Worked Examples
Worked Example: Compare a Rational Term with a p-Series
Determine whether \(\sum_{n=1}^{\infty}\frac{n}{n^3+1}\) converges.
First, the terms are positive because \(n>0\) and \(n^3+1>0\) for \(n\ge1\). To compare with \(1/n^2\), verify the inequality by multiplying by the positive denominators:
The last inequality is true for every \(n\ge1\). Therefore
The comparison series \(\sum_{n=1}^{\infty}1/n^2\) is a p-series with \(p=2>1\), so it converges. The Direct Comparison Test proves that \(\sum_{n=1}^{\infty}n/(n^3+1)\) converges.
Check the inequality at the first term: \(1/(1^3+1)=1/2\), while \(1/1^2=1\), so \(1/2\le1\). The general algebra above verifies the bound for every term, not just this check.
Worked Example: Compare with a Geometric Series
Determine whether \(\sum_{n=1}^{\infty}\frac{1}{3^n+2}\) converges.
Both the given terms and \(1/3^n\) are positive. Since \(3^n+2\ge3^n\), taking reciprocals of these positive quantities reverses the inequality:
The comparison series is geometric with first term \(1/3\) and ratio \(r=1/3\). Because \(|r|<1\), it converges, and its sum is
The given series therefore converges by direct comparison. As a check on the first term, \(1/(3+2)=1/5\le1/3\), in agreement with the established bound.
Worked Example: Bound a Radical Expression
Determine whether \(\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^4+n}+n^2}\) converges.
For \(n\ge1\), \(n^4+n\ge n^4\). Both sides are nonnegative, so taking square roots gives \(\sqrt{n^4+n}\ge\sqrt{n^4}=n^2\). Hence the denominator is at least \(2n^2\), and all quantities are positive:
The comparison series \(\sum_{n=1}^{\infty}1/(2n^2)\) is one-half of the convergent p-series \(\sum 1/n^2\). Thus the Direct Comparison Test proves convergence.
For a check at \(n=1\), the given term is \(1/(\sqrt2+1)\), and the bound is \(1/2\). Since \(\sqrt2+1\ge2\), taking reciprocals gives \(1/(\sqrt2+1)\le1/2\), as required.
Finding a Useful Comparison
A good comparison captures the size of a term without requiring an exact evaluation of the original series. When choosing one, focus on a bound that is both true and useful: the bound must go in the correct direction, and the comparison series must be known to converge.
Verify that the terms of both series are nonnegative from some point onward.
Look for a geometric term or a constant multiple of \(1/n^p\) with \(p>1\).
Use positive denominators or other valid algebra to establish \(0\le a_n\le b_n\).
State why the comparison series converges, then conclude that the original series converges.
For rational expressions, it is often effective to compare the highest powers in the numerator and denominator. But an informal observation about dominant powers is not itself an inequality. You still need to show the bound holds for the relevant indices.
A Common Mistake: Using the Wrong Inequality
Direct comparison for convergence requires an upper bound by a convergent series. For example, consider trying to prove convergence of \(\sum 1/n\) by writing
This inequality is false when \(n>1\). At \(n=2\), it would assert \(1/2\le1/4\), which is incorrect. In fact, for \(n\ge2\), the inequality goes the other way: \(1/n^2\le1/n\). It does not provide the required upper bound by the convergent p-series.
Another error is to find a true upper bound but forget to check whether its series converges. An upper bound by a divergent series does not meet the hypotheses of the Direct Comparison Test. Always verify both parts: the term inequality and convergence of the comparison series.
Tail Estimates and the Meaning of the Bound
Comparison also controls how much of the series remains after a chosen index. If \(0\le a_n\le b_n\) for \(n\ge N\), then for any \(M\ge N\),
This follows first for finite sums by adding the termwise inequalities, and then for the infinite tails by taking limits of those partial sums. Thus, a convergent comparison series can provide more than a yes-or-no convergence result: its tail can bound the tail of the original series.
The central logic remains the same throughout: nonnegative terms make partial sums increase, and a convergent upper comparison keeps those partial sums bounded. The direction of the inequality and the convergence of the benchmark are both essential.
Check Your Understanding
For each question, identify the inequality or condition that makes the comparison valid.
- To compare \(\sum_{n=1}^{\infty}\frac{n}{n^3+4}\) with \(\sum 1/n^2\), what inequality should you verify?
- Why does \(0\le a_n\le b_n\), together with convergence of \(\sum b_n\), bound the partial sums of \(\sum a_n\)?
- Does comparison with \(\sum 1/n\) establish convergence by the Direct Comparison Test? Explain what hypothesis is missing.
- If \(a_n\le b_n\) only for \(n\ge5\), why can the first four terms be set aside?
- For \(\sum_{n=1}^{\infty}1/(5^n+1)\), give a convergent geometric series that can serve as an upper comparison.
