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.

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

A particular case of Dirichlet's test is the more commonly used alternating series test for the case

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]

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.

• 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.

• PlanetMath.org