Dirichlet's test

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

• ${\displaystyle \lim _{n\rightarrow \infty }a_{n}=0}$

• ${\displaystyle \left|\sum _{n=1}^{N}b_{n}\right|\leq M}$ for every positive integer N

where M is some constant, then the series

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

converges.

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

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

We have, for each k, ${\displaystyle |B_{k}(a_{k}-a_{k+1})|\leq M|a_{k}-a_{k+1}|}$. But, 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, ${\displaystyle \sum _{k=1}^{\infty }M(a_{k}-a_{k+1})}$ converges. And, 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, ${\displaystyle \sum _{k=1}^{\infty }M(a_{k}-a_{k+1})}$ converges.

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

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 ${\displaystyle \sum _{n=1}^{\infty }a_{n}\sin n}$ converges whenever ${\displaystyle \{a_{n}\}}$ is a decreasing sequence that tends to zero.

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 monotonically decreasing non-negative function, then the integral of fg is a convergent improper integral.

• 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.

• Proof at PlanetMath.org