Dirichlet's test

Short description: Test for series convergence

In mathematics, Dirichlet's test is a method of testing for the convergence of a series. It is named after its author Peter Gustav Lejeune Dirichlet, and was published posthumously in the Journal de Mathématiques Pures et Appliquées in 1862.

Statement

The test states that if [math]\displaystyle{ (a_n) }[/math] is a sequence of real numbers and [math]\displaystyle{ (b_n) }[/math] a sequence of complex numbers satisfying

• [math]\displaystyle{ (a_n) }[/math] is monotonic
• [math]\displaystyle{ \lim_{n \to \infty}a_n = 0 }[/math]
• [math]\displaystyle{ \left|\sum_{n=1}^{N}b_n\right| \leq M }[/math] for every positive integer N

where M is some constant, then the series

[math]\displaystyle{ \sum_{n=1}^{\infty} a_n b_n }[/math]

converges.

Proof

Let [math]\displaystyle{ S_n = \sum_{k=1}^n a_k b_k }[/math] and [math]\displaystyle{ B_n = \sum_{k=1}^n b_k }[/math].

From summation by parts, we have that [math]\displaystyle{ S_n = a_n B_n + \sum_{k=1}^{n-1} B_k (a_k - a_{k+1}) }[/math]. Since [math]\displaystyle{ B_n }[/math] is bounded by M and [math]\displaystyle{ a_n \to 0 }[/math], the first of these terms approaches zero, [math]\displaystyle{ a_n B_n \to 0 }[/math] as [math]\displaystyle{ n\to\infty }[/math].

We have, for each k, [math]\displaystyle{ |B_k (a_k - a_{k+1})| \leq M|a_k - a_{k+1}| }[/math].

Since [math]\displaystyle{ (a_n) }[/math] is monotone, it is either decreasing or increasing:

• If [math]\displaystyle{ (a_n) }[/math] is decreasing, [math]\displaystyle{ \sum_{k=1}^n M|a_k - a_{k+1}| = \sum_{k=1}^n M(a_k - a_{k+1}) = M\sum_{k=1}^n (a_k - a_{k+1}), }[/math] which is a telescoping sum that equals [math]\displaystyle{ M(a_1 - a_{n+1}) }[/math] and therefore approaches [math]\displaystyle{ Ma_1 }[/math] as [math]\displaystyle{ n \to \infty }[/math]. Thus, [math]\displaystyle{ \sum_{k=1}^\infty M(a_k - a_{k+1}) }[/math] converges.
• If [math]\displaystyle{ (a_n) }[/math] is increasing, [math]\displaystyle{ \sum_{k=1}^n M|a_k - a_{k+1}| = -\sum_{k=1}^n M(a_k - a_{k+1}) = -M\sum_{k=1}^n (a_k - a_{k+1}), }[/math] which is again a telescoping sum that equals [math]\displaystyle{ -M(a_1 - a_{n+1}) }[/math] and therefore approaches [math]\displaystyle{ -Ma_1 }[/math] as [math]\displaystyle{ n\to\infty }[/math]. Thus, again, [math]\displaystyle{ \sum_{k=1}^\infty M(a_k - a_{k+1}) }[/math] converges.

So, the series [math]\displaystyle{ \sum_{k=1}^\infty B_k(a_k - a_{k+1}) }[/math] converges, by the absolute convergence test. Hence [math]\displaystyle{ S_n }[/math] converges.

Applications

A particular case of Dirichlet's test is the more commonly used alternating series test for the case [math]\displaystyle{ b_n = (-1)^n \Longrightarrow\left|\sum_{n=1}^N b_n\right| \leq 1. }[/math]

Another corollary is that [math]\displaystyle{ \sum_{n=1}^\infty a_n \sin n }[/math] converges whenever [math]\displaystyle{ (a_n) }[/math] is a decreasing sequence that tends to zero. To see that [math]\displaystyle{ \sum_{n=1}^N \sin n }[/math] is bounded, we can use the summation formula [math]\displaystyle{ \sum_{n=1}^N\sin n=\sum_{n=1}^N\frac{e^{in}-e^{-in}}{2i}=\frac{\sum_{n=1}^N (e^{i})^n-\sum_{n=1}^N (e^{-i})^n}{2i}=\frac{\sin 1 +\sin N-\sin (N+1)}{2- 2\cos 1}. }[/math]

Improper integrals

An analogous statement for convergence of improper integrals is proven using integration by parts. If the integral of a function f is uniformly bounded over all intervals, and g is a non-negative monotonically decreasing function, then the integral of fg is a convergent improper integral.

References

• Hardy, G. H., A Course of Pure Mathematics, Ninth edition, Cambridge University Press, 1946. (pp. 379–380).
• Voxman, William L., Advanced Calculus: An Introduction to Modern Analysis, Marcel Dekker, Inc., New York, 1981. (§8.B.13–15) ISBN:0-8247-6949-X.

External links

• PlanetMath.org