Direct comparison test

For series

In calculus, the comparison test for series typically consists of a pair of statements about infinite series with non-negative (real-valued) terms:

• If the infinite series ${\displaystyle \sum b_n}$ converges and ${\displaystyle 0\le a_n\le b_n}$ for all sufficiently large n (that is, for all ${\displaystyle n>N}$ for some fixed value N), then the infinite series ${\displaystyle \sum a_n}$ also converges.
• If the infinite series ${\displaystyle \sum b_n}$ diverges and ${\displaystyle 0\le b_n\le a_n}$ for all sufficiently large n, then the infinite series ${\displaystyle \sum a_n}$ also diverges.

Note that the series having larger terms is sometimes said to dominate (or eventually dominate) the series with smaller terms.

Alternatively, the test may be stated in terms of absolute convergence, in which case it also applies to series with complex terms:

• If the infinite series ${\displaystyle \sum b_n}$ is absolutely convergent and ${\displaystyle |a_n|\le |b_n|}$ for all sufficiently large n, then the infinite series ${\displaystyle \sum a_n}$ is also absolutely convergent.
• If the infinite series ${\displaystyle \sum b_n}$ is not absolutely convergent and ${\displaystyle |b_n|\le |a_n|}$ for all sufficiently large n, then the infinite series ${\displaystyle \sum a_n}$ is also not absolutely convergent.

Note that in this last statement, the series ${\displaystyle \sum a_n}$ could still be conditionally convergent; for real-valued series, this could happen if the a_n are not all nonnegative.

The second pair of statements are equivalent to the first in the case of real-valued series because ${\displaystyle \sum c_n}$ converges absolutely if and only if ${\displaystyle \sum |c_n|}$, a series with nonnegative terms, converges.

Proof

The proofs of all the statements given above are similar. Here is a proof of the third statement.

Let ${\displaystyle \sum a_n}$ and ${\displaystyle \sum b_n}$ be infinite series such that ${\displaystyle \sum b_n}$ converges absolutely (thus ${\displaystyle \sum |b_n|}$ converges), and without loss of generality assume that ${\displaystyle |a_n|\le |b_n|}$ for all positive integers n. Consider the partial sums

${\displaystyle S_n=|a_1|+|a_2|+\dots +|a_n|,T_n=|b_1|+|b_2|+\dots +|b_n|.}$

Since ${\displaystyle \sum b_n}$ converges absolutely, ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{T}_n=T}$ for some real number T. For all n,

${\displaystyle 0\le S_n=|a_1|+|a_2|+\dots +|a_n|\le |a_1|+\dots +|a_n|+|b_{n+1}|+\dots =S_n+(T-T_n)\le T.}$

${\displaystyle S_n}$ is a nondecreasing sequence and ${\displaystyle S_n+(T-T_n)}$ is nonincreasing. Given ${\displaystyle m,n>N}$ then both ${\displaystyle S_n,S_m}$ belong to the interval ${\displaystyle [S_N,S_N+(T-T_N)]}$, whose length ${\displaystyle T-T_N}$ decreases to zero as ${\displaystyle N}$ goes to infinity. This shows that ${\displaystyle (S_n{)}_{n=1,2,\dots }}$ is a Cauchy sequence, and so must converge to a limit. Therefore, ${\displaystyle \sum a_n}$ is absolutely convergent.