Limit comparison test

Part of a series of articles about

Definitions
Derivative (generalizations) Differential infinitesimal of a function total
Concepts
Differentiation notation Second derivative Third derivative Change of variables Implicit differentiation Related rates Taylor's theorem
Rules and identities
Sum Product Chain Power Quotient General Leibniz Faà di Bruno's formula

Integral

Definitions
Lists of integrals
Antiderivative Integral (improper) Riemann integral Lebesgue integration Contour integration
Integration by
Parts Discs Cylindrical shells Substitution (trigonometric) Partial fractions Order Reduction formulae

Series

Convergence tests
Geometric (arithmetico-geometric) Harmonic Alternating Power Binomial Taylor
Summand limit (term test) Ratio Root Integral Direct comparison Limit comparison Alternating series Cauchy condensation Dirichlet Abel

Vector

Theorems
Gradient Divergence Curl Laplacian Directional derivative Identities
Divergence Gradient Green's Kelvin–Stokes

Multivariable

Formalisms
Matrix Tensor Exterior Geometric
Definitions
Partial derivative Multiple integral Line integral Surface integral Volume integral Jacobian Hessian matrix

Specialized

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 $\Sigma_n a_n$ and $\Sigma_n b_n$ with $a_n, b_n \geq 0$ 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 all $\varepsilon >0$ there is a positive integer $n_{0}$ such that for all $n \geq n_0$ we have that $\left| \frac{a_n}{b_n} - c \right| < \varepsilon$ , or equivalently

- \varepsilon < \frac{a_n}{b_n} - c < \varepsilon

c - \varepsilon < \frac{a_n}{b_n} < c + \varepsilon

(c - \varepsilon)b_n < a_n < (c + \varepsilon)b_n

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

Similarly $a_n < (c + \varepsilon)b_n$ , so if $\sum_n b_n$ converges, 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 $\sum_{n=1}^{\infty} \frac{1}{n^2 + 2n}$ converges. For this we compare with the convergent series $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ .

As $\lim_{n \to \infty} \frac{1}{n^2 + 2n} \frac{n^2}{1} = 1 > 0$ 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 $\limsup _{{n\to \infty }}{\frac {a_{n}}{b_{n}}}=c$ with $0\leq c<\infty$ and $\Sigma_n b_n$ converges, necessarily $\Sigma_n a_n$ 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, $\limsup _{{n\to \infty }}{\frac {a_{n}}{b_{n}}}=\limsup _{{n\to \infty }}(1-(-1)^{n})=2\in [0,\infty )$ 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 $\Sigma _{n}a_{n}$ diverges and $\Sigma _{n}b_{n}$ converges, then necessarily $\limsup _{{n\to \infty }}{\frac {a_{n}}{b_{n}}}=\infty$ , that is, $\liminf _{{n\to \infty }}{\frac {b_{n}}{a_{n}}}=0$ . 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 $D=\{z\in {\mathbb {C}}:|z|<1\}$ and have image of finite area. By Parseval's formula the area of the image of $f$ is $\sum _{{n=1}}^{{\infty }}n|a_{n}|^{2}$ . Moreover, $\sum _{{n=1}}^{{\infty }}1/n$ diverges. Therefore, by the converse of the comparison test, we have $\liminf _{{n\to \infty }}{\frac {n|a_{n}|^{2}}{1/n}}=\liminf _{{n\to \infty }}(n|a_{n}|)^{2}=0$ , that is, $\liminf _{{n\to \infty }}n|a_{n}|=0$ .

References

↑ Swokowski, Earl (1983), Calculus with analytic geometry (Alternate ed.), Prindle, Weber & Schmidt, p. 516, ISBN 0-87150-341-7

External links

Pauls Online Notes on Comparison Test

This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.

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