Proof of the Comparison Test

From Termwise Inequalities to Series Behavior

The Comparison Test turns inequalities between nonnegative terms into conclusions about convergence and divergence. Its mechanism is simple but important: if \(a_n\leq b_n\), then every partial sum of the \(a_n\)'s is at most the corresponding partial sum of the \(b_n\)'s. The proof then uses what is already known about increasing sequences of partial sums.

We will allow the termwise inequality to hold only for all sufficiently large \(n\). This is often the useful form in applications. The Theorem (Finite Changes Preserve Series Convergence) ensures that changing or omitting finitely many terms does not affect convergence. For divergence of nonnegative series, finitely many initial terms likewise cannot make an unbounded sequence of partial sums bounded.

A Partial-Sum Fact for Nonnegative Series

For a series with nonnegative terms, its partial sums are nondecreasing. The earlier Theorem (Bounded Increasing Partial Sums) says that such a series converges if its partial sums are bounded. There is a complementary observation: if those partial sums are unbounded, they tend to infinity, not merely away from one particular value. This makes the divergence part of comparison especially transparent.

Proposition (Unbounded Increasing Partial Sums Tend to Infinity): Let \((S_N)\) be a nondecreasing sequence of real numbers. If \((S_N)\) is unbounded above, then for every real number \(K\), there is an index \(N_0\) such that \(S_N>K\) for every \(N\geq N_0\). In particular, \(S_N\to+\infty\).

Proof. Fix any real number \(K\). Since \((S_N)\) is unbounded above, there is an index \(N_0\) such that \(S_{N_0}>K\). For every \(N\geq N_0\), monotonicity gives \(S_N\geq S_{N_0}>K\). This is exactly the definition of \(S_N\to+\infty\). \(\square\)

For a series with nonnegative terms, the partial sums are nondecreasing because \(S_{N+1}=S_N+a_{N+1}\geq S_N\). Thus, if such a series diverges, the earlier Theorem (Bounded Increasing Partial Sums) implies that its partial sums are unbounded; the proposition then says they tend to infinity. We will use this fact when a divergent series is bounded above by another series term by term.

The Comparison Test and Its Proof

Theorem (Comparison Test for Nonnegative Series): Let \(a_n\geq0\) and \(b_n\geq0\) for every \(n\). If \(a_n\leq b_n\) for all sufficiently large \(n\), then:

If \(\sum b_n\) converges, then \(\sum a_n\) converges.
If \(\sum a_n\) diverges, then \(\sum b_n\) diverges.

Equivalently, a convergent upper comparison proves convergence, and a divergent lower comparison proves divergence.

Proof. First suppose that \(a_n\leq b_n\) for every \(n\geq N_0\), and that \(\sum b_n\) converges. For \(N\geq N_0\), split the partial sum of the \(a_n\)'s into its initial terms and its tail:

\sum_{n=1}^{N}a_n = \sum_{n=1}^{N_0-1}a_n+\sum_{n=N_0}^{N}a_n \leq \sum_{n=1}^{N_0-1}a_n+\sum_{n=N_0}^{N}b_n.

The initial sum on the right is a fixed finite number. The remaining partial sums of \(\sum b_n\) are bounded above because that series converges. Therefore the partial sums of \(\sum a_n\) are bounded above. They are nondecreasing, since \(a_n\geq0\). By the Theorem (Bounded Increasing Partial Sums), \(\sum a_n\) converges.

For the second assertion, suppose that \(\sum a_n\) diverges. Its partial sums are nondecreasing, so the proposition above shows that they tend to infinity. Removing the finitely many terms before \(N_0\) does not change that conclusion: the tail partial sums

\sum_{n=N_0}^{N}a_n = \sum_{n=1}^{N}a_n-\sum_{n=1}^{N_0-1}a_n

also tend to infinity, because the subtracted quantity is fixed. For \(N\geq N_0\), termwise comparison now gives

\sum_{n=N_0}^{N}b_n \geq \sum_{n=N_0}^{N}a_n.

Hence the partial sums of \(\sum b_n\) are unbounded above. Since its terms are nonnegative, the earlier Theorem (Bounded Increasing Partial Sums) implies that \(\sum b_n\) diverges. This proves both assertions. \(\square\)

The two conclusions are not interchangeable. From \(a_n\leq b_n\), convergence of the larger series controls the smaller one. But divergence of the larger series does not force the smaller one to diverge. The other conclusion uses the same inequality in reverse: if the smaller series already diverges, then the larger series must diverge too.

Worked Applications

Worked Example: Bounding a Rational Series Above

Consider

\sum_{n=1}^{\infty}\frac{1}{n^2+3n}.

For every \(n\geq1\), \(n^2+3n\geq n^2>0\), so taking reciprocals gives

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

To verify convergence of the upper comparison, for \(n\geq2\) we have \(n^2\geq n(n-1)>0\), and therefore

0<\frac{1}{n^2}\leq\frac{1}{n(n-1)} =\frac{1}{n-1}-\frac{1}{n}.

The finite sum of the right-hand terms from \(n=2\) through \(n=N\) is \(1-1/N\), which is at most \(1\). Thus the partial sums of \(\sum_{n=2}^{\infty}1/n^2\) are bounded; they are also nondecreasing. The Theorem (Bounded Increasing Partial Sums) gives convergence of \(\sum 1/n^2\). The Comparison Test then proves convergence of the given series.

Worked Example: A Lower Bound That Forces Divergence

Consider the positive terms

a_n=\frac{2n+3}{n^2+5n}.

For every \(n\geq1\), the denominator \(n^2+5n\) is positive. We compare \(a_n\) with \(1/(3n)\). After multiplying by the positive quantity \(3n(n^2+5n)\), the desired inequality is equivalent to

3n(2n+3)\geq n^2+5n.

The difference between the left and right sides is \(5n^2+4n\), which is positive for \(n\geq1\). Consequently,

\frac{2n+3}{n^2+5n}\geq\frac{1}{3n}.

The harmonic series \(\sum 1/n\) diverges, so \(\sum 1/(3n)\) diverges by linearity of convergent series, or equivalently because its partial sums are one third of the harmonic partial sums. The series with terms \(a_n\) is bounded below by this divergent series. The divergence conclusion of the Comparison Test therefore gives

\sum_{n=1}^{\infty}\frac{2n+3}{n^2+5n} \quad\text{diverges.}

Worked Example: Why a Divergent Upper Bound Is Inconclusive

For every \(n\geq1\),

0<\frac{1}{2^n}\leq\frac{1}{n}.

Indeed, \(2^n\geq n\) for \(n\geq1\): it holds at \(n=1\), and if \(2^n\geq n\), then \(2^{n+1}\geq2n\geq n+1\). The upper comparison series \(\sum 1/n\) diverges, but the lower series \(\sum 1/2^n\) converges, since it is geometric with ratio \(1/2\). Thus an inequality placing a series below a divergent series does not determine its behavior. The comparison theorem requires a convergent upper bound to establish convergence.

Using Eventual Inequalities Carefully

In practice, an inequality may become true only after some index \(N_0\). The proof works because the first \(N_0-1\) terms contribute only a fixed finite amount to each partial sum. The tail inequality controls all later terms, and the Theorem (Finite Changes Preserve Series Convergence) permits us to disregard the finite initial segment when deciding convergence.

The direction of the inequality should be checked before applying the test. To prove convergence, seek a termwise upper bound whose series converges. To prove divergence, seek a termwise lower bound whose series diverges. A comparison in the opposite direction may be correct algebraically but insufficient for the desired conclusion.

Check nonnegativity.
Partial sums must be nondecreasing for the positive-series comparison argument.

Establish the inequality.
Verify its direction, and if necessary identify an index after which it holds.

Choose the comparison's known behavior.
Use a convergent upper series to prove convergence, or a divergent lower series to prove divergence.

Transfer the conclusion.
Apply the Comparison Test to the tail and account for the finitely many initial terms.

Takeaway: A termwise inequality controls the corresponding partial sums. For nonnegative series, bounded partial sums give convergence, while unbounded partial sums tend to infinity. The direction of the inequality determines which convergence conclusion can be transferred.

Check Your Understanding

Use partial sums and the direction of comparison to answer these questions.

If \(0\leq a_n\leq b_n\) eventually and \(\sum b_n\) converges, what can you conclude about \(\sum a_n\)?
If \(0\leq a_n\leq b_n\) eventually and \(\sum a_n\) diverges, what can you conclude about \(\sum b_n\)?
Why do nonnegative terms ensure that the partial sums form a nondecreasing sequence?
Why does an unbounded nondecreasing sequence tend to infinity?
Does \(a_n\leq b_n\) and divergence of \(\sum b_n\) imply divergence of \(\sum a_n\)? Explain the role of the inequality's direction.

What You'll Learn