Direct comparison test explained

In mathematics, the comparison test, sometimes called the direct comparison test to distinguish it from similar related tests (especially the limit comparison test), provides a way of deducing whether an infinite series or an improper integral converges or diverges by comparing the series or integral to one whose convergence properties are known.

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:[1]

\sumbn

converges and

0\lean\lebn

for all sufficiently large n (that is, for all

n>N

for some fixed value N), then the infinite series

\suman

also converges.

\sumbn

diverges and

0\lebn\lean

for all sufficiently large n, then the infinite series

\suman

also diverges.Note that the series having larger terms is sometimes said to dominate (or eventually dominate) the series with smaller terms.[2]

Alternatively, the test may be stated in terms of absolute convergence, in which case it also applies to series with complex terms:[3]

\sumbn

is absolutely convergent and

|an|\le|bn|

for all sufficiently large n, then the infinite series

\suman

is also absolutely convergent.

\sumbn

is not absolutely convergent and

|bn|\le|an|

for all sufficiently large n, then the infinite series

\suman

is also not absolutely convergent.Note that in this last statement, the series

\suman

could still be conditionally convergent; for real-valued series, this could happen if the an are not all nonnegative.

The second pair of statements are equivalent to the first in the case of real-valued series because

\sumcn

converges absolutely if and only if

\sum|cn|

, 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

\suman

and

\sumbn

be infinite series such that

\sumbn

converges absolutely (thus

\sum|bn|

converges), and without loss of generality assume that

|an|\le|bn|

for all positive integers n. Consider the partial sums

Sn=|a1|+|a2|+\ldots+|an|,Tn=|b1|+|b2|+\ldots+|bn|.

Since

\sumbn

converges absolutely,

\limn\toinftyTn=T

for some real number T. For all n,

0\leSn=|a1|+|a2|+\ldots+|an|\le|a1|+\ldots+|an|+|bn+1|+\ldots=Sn+(T-Tn)\leT.

Sn

is a nondecreasing sequence and

Sn+(T-Tn)

is nonincreasing.Given

m,n>N

then both

Sn,Sm

belong to the interval

[SN,SN+(T-TN)]

, whose length

T-TN

decreases to zero as

N

goes to infinity.This shows that

(Sn)n=1,2,\ldots

is a Cauchy sequence, and so must converge to a limit. Therefore,

\suman

is absolutely convergent.

For integrals

The comparison test for integrals may be stated as follows, assuming continuous real-valued functions f and g on

[a,b)

with b either

+infty

or a real number at which f and g each have a vertical asymptote:[4]
b
\int
a

g(x)dx

converges and

0\lef(x)\leg(x)

for

a\lex<b

, then the improper integral
b
\int
a

f(x)dx

also converges with
b
\int
a

f(x)dx\le

b
\int
a

g(x)dx.

b
\int
a

g(x)dx

diverges and

0\leg(x)\lef(x)

for

a\lex<b

, then the improper integral
b
\int
a

f(x)dx

also diverges.

Ratio comparison test

Another test for convergence of real-valued series, similar to both the direct comparison test above and the ratio test, is called the ratio comparison test:[5]

\sumbn

converges and

an>0

,

bn>0

, and
an+1
an

\le

bn+1
bn
for all sufficiently large n, then the infinite series

\suman

also converges.

\sumbn

diverges and

an>0

,

bn>0

, and
an+1
an

\ge

bn+1
bn
for all sufficiently large n, then the infinite series

\suman

also diverges.

See also

Notes

  1. Ayres & Mendelson (1999), p. 401.
  2. Munem & Foulis (1984), p. 662.
  3. Silverman (1975), p. 119.
  4. Buck (1965), p. 140.
  5. Buck (1965), p. 161.

References