Limit comparison test
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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 [math]\displaystyle{ \Sigma_n a_n }[/math] and [math]\displaystyle{ \Sigma_n b_n }[/math] with [math]\displaystyle{ a_n\geq 0, b_n \gt 0 }[/math] for all [math]\displaystyle{ n }[/math]. Then if [math]\displaystyle{ \lim_{n \to \infty} \frac{a_n}{b_n} = c }[/math] with [math]\displaystyle{ 0 \lt c \lt \infty }[/math], then either both series converge or both series diverge.[1]
Proof
Because [math]\displaystyle{ \lim_{n \to \infty} \frac{a_n}{b_n} = c }[/math] we know that for every [math]\displaystyle{ \varepsilon \gt 0 }[/math] there is a positive integer [math]\displaystyle{ n_0 }[/math] such that for all [math]\displaystyle{ n \geq n_0 }[/math] we have that [math]\displaystyle{ \left| \frac{a_n}{b_n} - c \right| \lt \varepsilon }[/math], or equivalently
- [math]\displaystyle{ - \varepsilon \lt \frac{a_n}{b_n} - c \lt \varepsilon }[/math]
- [math]\displaystyle{ c - \varepsilon \lt \frac{a_n}{b_n} \lt c + \varepsilon }[/math]
- [math]\displaystyle{ (c - \varepsilon)b_n \lt a_n \lt (c + \varepsilon)b_n }[/math]
As [math]\displaystyle{ c \gt 0 }[/math] we can choose [math]\displaystyle{ \varepsilon }[/math] to be sufficiently small such that [math]\displaystyle{ c-\varepsilon }[/math] is positive. So [math]\displaystyle{ b_n \lt \frac{1}{c-\varepsilon} a_n }[/math] and by the direct comparison test, if [math]\displaystyle{ \sum_n a_n }[/math] converges then so does [math]\displaystyle{ \sum_n b_n }[/math].
Similarly [math]\displaystyle{ a_n \lt (c + \varepsilon)b_n }[/math], so if [math]\displaystyle{ \sum_n a_n }[/math] diverges, again by the direct comparison test, so does [math]\displaystyle{ \sum_n b_n }[/math].
That is, both series converge or both series diverge.
Example
We want to determine if the series [math]\displaystyle{ \sum_{n=1}^{\infty} \frac{1}{n^2 + 2n} }[/math] converges. For this we compare it with the convergent series [math]\displaystyle{ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} }[/math]
As [math]\displaystyle{ \lim_{n \to \infty} \frac{1}{n^2 + 2n} \frac{n^2}{1} = 1 \gt 0 }[/math] we have that the original series also converges.
One-sided version
One can state a one-sided comparison test by using limit superior. Let [math]\displaystyle{ a_n, b_n \geq 0 }[/math] for all [math]\displaystyle{ n }[/math]. Then if [math]\displaystyle{ \limsup_{n \to \infty} \frac{a_n}{b_n} = c }[/math] with [math]\displaystyle{ 0 \leq c \lt \infty }[/math] and [math]\displaystyle{ \Sigma_n b_n }[/math] converges, necessarily [math]\displaystyle{ \Sigma_n a_n }[/math] converges.
Example
Let [math]\displaystyle{ a_n = \frac{1-(-1)^n}{n^2} }[/math] and [math]\displaystyle{ b_n = \frac{1}{n^2} }[/math] for all natural numbers [math]\displaystyle{ n }[/math]. Now [math]\displaystyle{ \lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty}(1-(-1)^n) }[/math] does not exist, so we cannot apply the standard comparison test. However, [math]\displaystyle{ \limsup_{n\to\infty} \frac{a_n}{b_n} = \limsup_{n\to\infty}(1-(-1)^n) =2\in [0,\infty) }[/math] and since [math]\displaystyle{ \sum_{n=1}^{\infty} \frac{1}{n^2} }[/math] converges, the one-sided comparison test implies that [math]\displaystyle{ \sum_{n=1}^{\infty}\frac{1-(-1)^n}{n^2} }[/math] converges.
Converse of the one-sided comparison test
Let [math]\displaystyle{ a_n, b_n \geq 0 }[/math] for all [math]\displaystyle{ n }[/math]. If [math]\displaystyle{ \Sigma_n a_n }[/math] diverges and [math]\displaystyle{ \Sigma_n b_n }[/math] converges, then necessarily [math]\displaystyle{ \limsup_{n\to\infty} \frac{a_n}{b_n}=\infty }[/math], that is, [math]\displaystyle{ \liminf_{n\to\infty} \frac{b_n}{a_n}= 0 }[/math]. The essential content here is that in some sense the numbers [math]\displaystyle{ a_n }[/math] are larger than the numbers [math]\displaystyle{ b_n }[/math].
Example
Let [math]\displaystyle{ f(z)=\sum_{n=0}^{\infty}a_nz^n }[/math] be analytic in the unit disc [math]\displaystyle{ D = \{ z\in\mathbb{C} : |z|\lt 1\} }[/math] and have image of finite area. By Parseval's formula the area of the image of [math]\displaystyle{ f }[/math] is proportional to [math]\displaystyle{ \sum_{n=1}^{\infty} n|a_n|^2 }[/math]. Moreover, [math]\displaystyle{ \sum_{n=1}^{\infty} 1/n }[/math] diverges. Therefore, by the converse of the comparison test, we have [math]\displaystyle{ \liminf_{n\to\infty} \frac{n|a_n|^2}{1/n}= \liminf_{n\to\infty} (n|a_n|)^2 = 0 }[/math], that is, [math]\displaystyle{ \liminf_{n\to\infty} n|a_n| = 0 }[/math].
See also
References
- ↑ Swokowski, Earl (1983), Calculus with analytic geometry (Alternate ed.), Prindle, Weber & Schmidt, p. 516, ISBN 0-87150-341-7, https://archive.org/details/calculuswithanal00swok/page/516
Further reading
- Rinaldo B. Schinazi: From Calculus to Analysis. Springer, 2011, ISBN 9780817682897, pp. 50
- Michele Longo and Vincenzo Valori: The Comparison Test: Not Just for Nonnegative Series. Mathematics Magazine, Vol. 79, No. 3 (Jun., 2006), pp. 205–210 (JSTOR)
- J. Marshall Ash: The Limit Comparison Test Needs Positivity. Mathematics Magazine, Vol. 85, No. 5 (December 2012), pp. 374–375 (JSTOR)
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