Dirichlet's test

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.^[1]

Statement

The test states that if ${\displaystyle (a_{n})}$ is a sequence of real numbers and ${\displaystyle (b_{n})}$ a sequence of complex numbers satisfying

• ${\displaystyle (a_{n})}$ is monotonic
• ${\displaystyle \lim _{n\to \infty }a_{n}=0}$
• ${\displaystyle \left|\sum _{n=1}^{N}b_{n}\right|\leq M}$ for every positive integer N

where M is some constant, then the series

$${\displaystyle \sum _{n=1}^{\infty }a_{n}b_{n}}$$

converges.

Proof

Let ${\textstyle S_{n}=\sum _{k=1}^{n}a_{k}b_{k}}$ and ${\textstyle B_{n}=\sum _{k=1}^{n}b_{k}}$.

From summation by parts, we have that ${\textstyle S_{n}=a_{n}B_{n}+\sum _{k=1}^{n-1}B_{k}(a_{k}-a_{k+1})}$. Since ${\displaystyle B_{n}}$ is bounded by M and ${\displaystyle a_{n}\to 0}$, the first of these terms approaches zero, ${\displaystyle a_{n}B_{n}\to 0}$ as ${\displaystyle n\to \infty }$.

We have, for each k, ${\displaystyle |B_{k}(a_{k}-a_{k+1})|\leq M|a_{k}-a_{k+1}|}$.

Since ${\displaystyle (a_{n})}$ is monotone, it is either decreasing or increasing:

• If ${\displaystyle (a_{n})}$ is decreasing, $${\displaystyle \sum _{k=1}^{n}M|a_{k}-a_{k+1}|=\sum _{k=1}^{n}M(a_{k}-a_{k+1})=M\sum _{k=1}^{n}(a_{k}-a_{k+1}),}$$which is a telescoping sum that equals ${\displaystyle M(a_{1}-a_{n+1})}$ and therefore approaches ${\displaystyle Ma_{1}}$ as ${\displaystyle n\to \infty }$. Thus, ${\textstyle \sum _{k=1}^{\infty }M(a_{k}-a_{k+1})}$ converges.
• If ${\displaystyle (a_{n})}$ is increasing, $${\displaystyle \sum _{k=1}^{n}M|a_{k}-a_{k+1}|=-\sum _{k=1}^{n}M(a_{k}-a_{k+1})=-M\sum _{k=1}^{n}(a_{k}-a_{k+1}),}$$which is again a telescoping sum that equals ${\displaystyle -M(a_{1}-a_{n+1})}$ and therefore approaches ${\displaystyle -Ma_{1}}$ as ${\displaystyle n\to \infty }$. Thus, again, ${\textstyle \sum _{k=1}^{\infty }M(a_{k}-a_{k+1})}$ converges.

So, the series ${\textstyle \sum _{k=1}^{\infty }B_{k}(a_{k}-a_{k+1})}$ converges, by the absolute convergence test. Hence ${\displaystyle S_{n}}$ converges.

Applications

A particular case of Dirichlet's test is the more commonly used alternating series test for the case $${\displaystyle b_{n}=(-1)^{n}\Longrightarrow \left|\sum _{n=1}^{N}b_{n}\right|\leq 1.}$$

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

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.

Notes

1. Démonstration d’un théorème d’Abel. Journal de mathématiques pures et appliquées 2nd series, tome 7 (1862), pp. 253–255 Archived 2011-07-21 at the Wayback Machine. See also [1].
2. "Where does the sum of $\sin(n)$ formula come from?".

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