Comparison Test for Series¶

Proposition¶

Suppose we have two series $\displaystyle\sum_{n=1}^\infty a_n$ and $\displaystyle\sum_{n=1}^\infty b_n$ with $a_n, b_n \geq 0$ for all $n\text{,}$ and $a_n \leq b_n$ for all $n\text{.}$ Then:

If $\displaystyle\sum_{n=1}^\infty b_n$ is convergent, then $\displaystyle\sum_{n=1}^\infty a_n$ is also convergent.
If $\displaystyle\sum_{n=1}^\infty a_n$ is divergent, then $\displaystyle\sum_{n=1}^\infty b_n$ is also divergent.

Interpretation¶

The comparison test decides the convergence of a series with nonnegative terms by comparing it, term by term, with a series whose behaviour is known: a series below a convergent one converges, and a series above a divergent one diverges.

Two sets of bars, the larger terms b n = 1/n² and the smaller terms a n = 1/(n(n+1)), with each a n at most b n.

The bars show $b_n=1/n^2$ and the smaller terms $a_n=1/(n(n+1))$ for $n=1,\ldots,9$ $1, \dots, 9$ ; for example, $a_1=\tfrac{1}{2}\le b_1=1$ $1$ . Since $a_n\le b_n$ for every $n$ , each partial sum of $\displaystyle \sum a_n$ is at most the corresponding partial sum of $\displaystyle \sum b_n$ , and hence at most $\displaystyle \sum b_n<\infty$ $\infty$ .

That bound is the whole proof: because $a_n\ge0$ $0$ , the partial sums of $\displaystyle \sum a_n$ increase, and an increasing sequence that is bounded above converges by the monotone convergence theorem. This is why nonnegativity is needed: $a_n=-1\le b_n=0$ $0$ , yet $\displaystyle \sum b_n$ converges and $\displaystyle \sum a_n$ diverges. The inequality $a_n\le b_n$ only needs to hold from some index on, because finitely many terms do not affect convergence.

A common slip is to compare in the wrong direction. A series below a divergent one may converge or diverge: $\tfrac{1}{n^2}\le\tfrac{1}{n}$ and $\displaystyle \sum 1/n^2$ converges, while $\tfrac{1}{2n}\le\tfrac{1}{n}$ and $\displaystyle \sum 1/(2n)$ diverges. In such cases the test is silent, and a different comparison is needed. The usual benchmarks are the geometric series and the harmonic series. The same idea returns for integrals in the comparison test for improper integrals.

Key Results¶

Comparison requires all terms to be nonnegative.
A series can be shown convergent by comparison with a larger convergent series.
Divergence can be shown by comparison with a smaller divergent series.

Examples¶

$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$ converges, so $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2+1}$ also converges.
Since $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$ diverges, so does $\displaystyle \sum_{n=1}^\infty \frac{1}{\sqrt{n}}\text{.}$

Prerequisites¶

Infinite Series

Sources¶

Math I: Calculus · Lecture 10

Aliases: direct comparison test Symbols: infinity, square root, sum

Last updated: 20 October 2026 03:02