Proof of the Direct Comparison Convergence Rule

Why an Upper Bound on Terms Can Control a Series

A series is defined through its finite partial sums. For nonnegative terms, those partial sums can only stay the same or increase. If we can also show that every partial sum is bounded above by a fixed finite number, then the partial sums cannot grow without limit. The bounded monotone sequence theorem then guarantees that they approach a finite limit.

The direct comparison convergence rule makes this idea precise. It says that if the terms of one nonnegative series are eventually no larger than the terms of a convergent nonnegative series, the first series converges as well. The key is not merely that each term is small: it is that the accumulated sums inherit a finite upper bound.

Definition: For a sequence of real numbers \(a_n\), its \(m\)th partial sum is \(S_m=\sum_{n=1}^{m}a_n\). The series \(\sum_{n=1}^{\infty}a_n\) converges if the sequence of partial sums \((S_m)\) converges to a finite real number.

In this tutorial, the terms being compared are nonnegative. Thus the partial sums are nondecreasing. The proof will also allow the inequality to hold only after some index: changing or leaving out finitely many initial terms cannot change whether the sequence of partial sums has a finite limit.

The Bounded Monotone Sequence Theorem

The central sequence result used in the proof is the bounded monotone sequence theorem: every nondecreasing sequence of real numbers that is bounded above converges to a finite real number. Here, “nondecreasing” means that each term is at least as large as the preceding term.

For series with nonnegative terms, the partial sums have exactly this monotonicity. If \(a_n\geq 0\), then

S_{m+1}=S_m+a_{m+1}\geq S_m.

Consequently, to prove convergence of such a series, it is enough to find a finite upper bound for all its partial sums. Direct comparison supplies that bound by relating the partial sums to those of a known convergent series.

Theorem: Suppose \(a_n\geq 0\) and \(b_n\geq 0\) for every \(n\in\mathbb{N}\). If \(\sum_{n=1}^{\infty}b_n\) converges and there is an integer \(N\in\mathbb{N}\) such that \(a_n\leq b_n\) for every \(n\geq N\), then \(\sum_{n=1}^{\infty}a_n\) converges.

The assumptions are important. Nonnegativity ensures that the partial sums of the series being proved convergent are nondecreasing. Convergence of the comparison series provides a finite upper bound. The inequality is required only from some index onward; the terms before that index are finite in number and have finite values.

Proof by Comparing Partial Sums

Let \(S_m=\sum_{n=1}^{m}a_n\). We separate the initial terms, before the comparison is known to hold, from the remaining terms. For \(m\geq N\), write

S_m = \sum_{n=1}^{N-1}a_n+\sum_{n=N}^{m}a_n.

Termwise comparison bounds the second sum by a partial sum of the convergent comparison series. Since every \(b_n\) is nonnegative, that partial sum is no greater than the entire sum of the comparison series. Therefore

\sum_{n=N}^{m}a_n \leq \sum_{n=N}^{m}b_n \leq \sum_{n=N}^{\infty}b_n.

The last quantity is finite: it is a tail of a convergent series with nonnegative terms. The initial sum \(\sum_{n=1}^{N-1}a_n\) is also finite because it contains only finitely many real numbers. Thus every \(S_m\) with \(m\geq N\) is bounded above by the same finite number:

S_m \leq \sum_{n=1}^{N-1}a_n+\sum_{n=N}^{\infty}b_n.

Also, \(S_{m+1}\geq S_m\), since \(a_{m+1}\geq 0\). The partial sums are therefore nondecreasing and bounded above. By the bounded monotone sequence theorem, \((S_m)\) converges to a finite limit. That is precisely the definition of convergence of \(\sum_{n=1}^{\infty}a_n\), proving the result.

Use the eventual inequality.
For \(n\geq N\), add \(a_n\leq b_n\) over the finite range from \(N\) to \(m\).

Obtain a uniform upper bound.
The comparison tail is no larger than the finite sum of the convergent series; the finitely many initial \(a_n\) add only a fixed amount.

Apply monotone convergence.
Nonnegative \(a_n\) make the partial sums nondecreasing, so their finite upper bound implies that they converge.

Worked Examples

Worked Example: A Telescoping Upper Bound

Prove that \(\sum_{n=1}^{\infty}\frac{1}{n^2+3n}\) converges by comparison.

For every \(n\geq 1\), \(n^2+3n\geq n^2+n=n(n+1)>0\). Taking reciprocals of positive quantities reverses the inequality, so

0<\frac{1}{n^2+3n}\leq\frac{1}{n(n+1)}.

The comparison terms telescope because

\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.

For any positive integer \(m\), their partial sum is

\sum_{n=1}^{m}\frac{1}{n(n+1)} = \left(1-\frac12\right) +\left(\frac12-\frac13\right) +\cdots +\left(\frac1m-\frac{1}{m+1}\right) = 1-\frac{1}{m+1}<1.

These partial sums increase and are bounded above by \(1\), so the comparison series converges. The direct comparison theorem applies with the inequality holding for every \(n\), and the original series converges. As a check, the denominator inequality holds for all \(n\geq1\), and the displayed telescoping sum is indeed less than \(1\) for every \(m\).

Worked Example: Bounding by a Geometric Series

Show that \(\sum_{n=1}^{\infty}\frac{1}{2^n+n}\) converges.

For \(n\geq1\), the denominator satisfies \(2^n+n\geq 2^n>0\). Hence

0<\frac{1}{2^n+n}\leq\frac{1}{2^n}=\left(\frac12\right)^n.

The comparison series is geometric with first term \(1/2\) and common ratio \(1/2\). Since the absolute value of its common ratio is less than \(1\), it converges. More explicitly, for every positive integer \(m\),

\sum_{n=1}^{m}\left(\frac12\right)^n = \frac{\frac12\left(1-(\frac12)^m\right)}{1-\frac12} = 1-\frac{1}{2^m}<1.

Its partial sums are increasing and bounded above by \(1\); their limit is \(1\). The terms of the given series are nonnegative and no greater than the comparison terms for every \(n\), so the direct comparison theorem proves convergence. Substituting, for example, \(m=2\) into the comparison formula gives \(1-\frac14=\frac34\), matching \(\frac12+\frac14\); the finite-sum formula and the termwise inequality are consistent.

Worked Example: An Inequality That Holds Eventually

Define \(a_1=5\) and \(a_n=\frac{1}{n(n+1)}\) for \(n\geq2\). Define \(b_1=1\) and \(b_n=\frac{1}{n(n-1)}\) for \(n\geq2\). Prove that \(\sum_{n=1}^{\infty}a_n\) converges by comparison.

The inequality \(a_1\leq b_1\) fails, since \(5>1\). But for every \(n\geq2\), both denominators are positive and \(n(n+1)>n(n-1)\), so

0<a_n=\frac{1}{n(n+1)} < \frac{1}{n(n-1)}=b_n.

The comparison series converges. Indeed, for \(n\geq2\),

b_n = \frac{1}{n(n-1)} = \frac{1}{n-1}-\frac{1}{n}.

Thus, for \(m\geq2\),

\sum_{n=1}^{m}b_n = 1+\sum_{n=2}^{m}\left(\frac{1}{n-1}-\frac{1}{n}\right) = 1+1-\frac1m = 2-\frac1m.

These partial sums tend to \(2\), so \(\sum b_n\) converges. Although the termwise inequality fails at \(n=1\), it holds for all \(n\geq2\). The theorem therefore applies with \(N=2\), proving that \(\sum a_n\) converges. As a direct check on the effect of the exceptional initial term, for \(m\geq2\) its partial sums are

\sum_{n=1}^{m}a_n = 5+\sum_{n=2}^{m}\left(\frac1n-\frac{1}{n+1}\right) = 5+\frac12-\frac{1}{m+1}.

They tend to \(11/2\). The first term changes the value of the sum but does not prevent convergence.

Why Finitely Many Exceptions Are Harmless

The proof makes clear why an eventual inequality is sufficient. Once \(n\geq N\), the comparison controls every added term in the tail. The terms \(a_1,\ldots,a_{N-1}\) contribute only the fixed finite amount \(\sum_{n=1}^{N-1}a_n\). They cannot make the partial sums unbounded or change whether those partial sums have a finite limit.

This does not mean the first terms are irrelevant to the value of the sum. They can change it substantially, as the last example shows. The claim is only about convergence: adding or changing finitely many finite terms cannot turn a convergent series into a divergent one, or vice versa.

A Common Comparison Mistake

A comparison must place the series being tested below a series already known to converge. For instance, the inequality

\frac{1}{n}\leq\frac{1}{n^2}

is false for \(n\geq2\): at \(n=2\), it says \(\frac12\leq\frac14\). In fact, the inequality points the other way. Even if one notices that \(1/n^2\leq1/n\), this does not prove convergence of \(\sum 1/n^2\) using the harmonic series as an upper bound: the harmonic series is not a convergent comparison series.

For a valid upper bound, use \(\frac{1}{n^2}\leq\frac{1}{n(n-1)}\) for \(n\geq2\). The right-hand side forms a convergent telescoping series, since

\sum_{n=2}^{m}\frac{1}{n(n-1)} = \sum_{n=2}^{m}\left(\frac{1}{n-1}-\frac{1}{n}\right) = 1-\frac1m \leq1.

The corrected comparison has the needed direction and a convergent upper series. The direction and the convergence of the comparison series are both essential to this rule.

What the Proof Establishes

The direct comparison convergence rule is a result about accumulated sums, not a conclusion drawn from a few small terms. Its proof has three necessary ingredients: nonnegative terms make partial sums nondecreasing, an eventual upper comparison bounds those partial sums, and bounded monotone sequences converge. Without the nonnegativity condition, partial sums need not be nondecreasing; without a finite upper bound, monotonicity alone does not guarantee a finite limit.

Takeaway: If \(0\leq a_n\leq b_n\) for all sufficiently large \(n\), and the nonnegative comparison series \(\sum b_n\) converges, then the partial sums of \(\sum a_n\) are nondecreasing and bounded above. Therefore \(\sum a_n\) converges.

Check Your Understanding

Answer each question using the partial-sum argument from the proof.

Why are the partial sums of a series with nonnegative terms nondecreasing?
In the direct comparison convergence theorem, why is it enough for the inequality to hold only for \(n\geq N\)?
Which two facts about the partial sums allow the bounded monotone sequence theorem to be applied?
Suppose \(0\leq a_n\leq b_n\) eventually, but \(\sum b_n\) is not known to converge. Which part of the proof is missing?
For \(n\geq2\), verify the inequality \(\frac{1}{n^2}\leq\frac{1}{n(n-1)}\), and state why the right-hand comparison series converges.