Limit comparison test

Template:Short description {{#invoke:sidebar|collapsible | class = plainlist | titlestyle = padding-bottom:0.25em; | pretitle = Part of a series of articles about | title = Calculus | image = $\int_{a}^{b} f^{'} (t) d t = f (b) - f (a)$ | listtitlestyle = text-align:center; | liststyle = border-top:1px solid #aaa;padding-top:0.15em;border-bottom:1px solid #aaa; | expanded = series | abovestyle = padding:0.15em 0.25em 0.3em;font-weight:normal; | above =

| list2name = differential | list2titlestyle = display:block;margin-top:0.65em; | list2title = Template:Bigger | list2 ={{#invoke:sidebar|sidebar|child=yes

 |contentclass=hlist
 | heading1 = Definitions
 | content1 =

 | heading2 = Concepts
 | content2 =

 | heading3 = Rules and identities
 | content3 =

}}

| list3name = integral | list3title = Template:Bigger | list3 ={{#invoke:sidebar|sidebar|child=yes

 |contentclass=hlist
 | content1 =

| heading2 = Definitions

 | content2 =

 | heading3 = Integration by
 | content3 =

}}

| list4name = series | list4title = Template:Bigger | list4 ={{#invoke:sidebar|sidebar|child=yes

 |contentclass=hlist
 | content1 =

 | heading2 = Convergence tests
 | content2 =

}}

| list5name = vector | list5title = Template:Bigger | list5 ={{#invoke:sidebar|sidebar|child=yes

 |contentclass=hlist
 | content1 =

 | heading2 = Theorems
 | content2 =

}}

| list6name = multivariable | list6title = Template:Bigger | list6 ={{#invoke:sidebar|sidebar|child=yes

 |contentclass=hlist
 | heading1 = Formalisms
 | content1 =

 | heading2 = Definitions
 | content2 =

}}

| list7name = advanced | list7title = Template:Bigger | list7 ={{#invoke:sidebar|sidebar|child=yes

 |contentclass=hlist
 | content1 =

}}

| list8name = specialized | list8title = Template:Bigger | list8 =

| list9name = miscellanea | list9title = Template:Bigger | list9 =

}}

In mathematics, the limit comparison test (LCT) (in contrast with the related direct comparison test) is a method of testing for the convergence of an infinite series.

Statement

Suppose that we have two series $Σ_{n} a_{n}$ and $Σ_{n} b_{n}$ with $a_{n} \geq 0, b_{n} > 0$ for all $n$ . Then if $\lim_{n \to \infty} \frac{a_{n}}{b_{n}} = c$ with $0 < c < \infty$ , then either both series converge or both series diverge.^[1]

Proof

Because $\lim_{n \to \infty} \frac{a_{n}}{b_{n}} = c$ we know that for every $ε > 0$ there is a positive integer $n_{0}$ such that for all $n \geq n_{0}$ we have that $| \frac{a_{n}}{b_{n}} - c | < ε$ , or equivalently

- ε < \frac{a_{n}}{b_{n}} - c < ε

c - ε < \frac{a_{n}}{b_{n}} < c + ε

(c - ε) b_{n} < a_{n} < (c + ε) b_{n}

As $c > 0$ we can choose $ε$ to be sufficiently small such that $c - ε$ is positive. So $b_{n} < \frac{1}{c - ε} a_{n}$ and by the direct comparison test, if $\sum_{n} a_{n}$ converges then so does $\sum_{n} b_{n}$ .

Similarly $a_{n} < (c + ε) b_{n}$ , so if $\sum_{n} b_{n}$ diverges, again by the direct comparison test, so does $\sum_{n} a_{n}$ .

That is, both series converge or both series diverge.

Example

We want to determine if the series $\sum_{n = 1}^{\infty} \frac{1}{n^{2} + 2 n}$ converges. For this we compare it with the convergent series $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6}$

As $\lim_{n \to \infty} \frac{1}{n^{2} + 2 n} \frac{n^{2}}{1} = 1 > 0$ we have that the original series also converges.

One-sided version

One can state a one-sided comparison test by using limit superior. Let $a_{n}, b_{n} \geq 0$ for all $n$ . Then if $\underset{n \to \infty}{lim sup} \frac{a_{n}}{b_{n}} = c$ with $0 \leq c < \infty$ and $Σ_{n} b_{n}$ converges, necessarily $Σ_{n} a_{n}$ converges.

Example

Let $a_{n} = \frac{1 - (- 1)^{n}}{n^{2}}$ and $b_{n} = \frac{1}{n^{2}}$ for all natural numbers $n$ . Now $\lim_{n \to \infty} \frac{a_{n}}{b_{n}} = \lim_{n \to \infty} (1 - (- 1)^{n})$ does not exist, so we cannot apply the standard comparison test. However, $\underset{n \to \infty}{lim sup} \frac{a_{n}}{b_{n}} = \underset{n \to \infty}{lim sup} (1 - (- 1)^{n}) = 2 \in [0, \infty)$ and since $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ converges, the one-sided comparison test implies that $\sum_{n = 1}^{\infty} \frac{1 - (- 1)^{n}}{n^{2}}$ converges.

Converse of the one-sided comparison test

Let $a_{n}, b_{n} \geq 0$ for all $n$ . If $Σ_{n} a_{n}$ diverges and $Σ_{n} b_{n}$ converges, then necessarily $\underset{n \to \infty}{lim sup} \frac{a_{n}}{b_{n}} = \infty$ , that is, $\underset{n \to \infty}{lim inf} \frac{b_{n}}{a_{n}} = 0$ . The essential content here is that in some sense the numbers $a_{n}$ are larger than the numbers $b_{n}$ .

Example

Let $f (z) = \sum_{n = 0}^{\infty} a_{n} z^{n}$ be analytic in the unit disc $D = {z \in ℂ : | z | < 1}$ and have image of finite area. By Parseval's formula the area of the image of $f$ is proportional to $\sum_{n = 1}^{\infty} n | a_{n} |^{2}$ . Moreover, $\sum_{n = 1}^{\infty} 1 / n$ diverges. Therefore, by the converse of the comparison test, we have $\underset{n \to \infty}{lim inf} \frac{n | a_{n} |^{2}}{1 / n} = \underset{n \to \infty}{lim inf} (n | a_{n} |)^{2} = 0$ , that is, $\underset{n \to \infty}{lim inf} n | a_{n} | = 0$ .

References

Template:Reflist

External links

Pauls Online Notes on Comparison Test

Template:Calculus topics

↑ Template:Citation

[1] Template:Citation

[1]

Limit comparison test

Contents

Statement

Proof

Example

One-sided version

Example

Converse of the one-sided comparison test

Example

See also

References

Further reading

External links

Navigation menu

Limit comparison test

Statement

Proof

Example

One-sided version

Example

Converse of the one-sided comparison test

Example

See also

References

Further reading

External links

Navigation menu

Search