The test states that if ${\displaystyle (a_{n})}$  is a monotonic sequence of real numbers with ${\textstyle \lim _{n\to \infty }a_{n}=0}$  and ${\displaystyle (b_{n})}$  is a sequence of real numbers or complex numbers with bounded partial sums, then the series

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

converges.^[2][3][4]

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 the magnitudes of the partial sums ${\displaystyle B_{n}}$  are bounded by some M and ${\displaystyle a_{n}\to 0}$  as ${\displaystyle n\to \infty }$ , the first of these terms approaches zero: ${\displaystyle |a_{n}B_{n}|\leq |a_{n}M|\to 0}$  as ${\displaystyle n\to \infty }$ .

Furthermore, 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 direct comparison test to ${\textstyle \sum _{k=1}^{\infty }M(a_{k}-a_{k+1})}$ . Hence ${\displaystyle S_{n}}$  converges.^[2][4]

A particular case of Dirichlet's test is the more commonly used alternating series test for the case^[2][5] $${\displaystyle b_{n}=(-1)^{n}\Longrightarrow \left|\sum _{n=1}^{N}b_{n}\right|\leq 1.}$$ 

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^[6] $${\displaystyle \sum _{n=1}^{N}\sin n=\sum _{n=1}^{N}{\frac {e^{in}-e^{-in}}{2i}}={\frac {\sum _{n=1}^{N}(e^{i})^{n}-\sum _{n=1}^{N}(e^{-i})^{n}}{2i}}={\frac {\sin 1+\sin N-\sin(N+1)}{2-2\cos 1}}.}$$ 

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.

• Apostol, Tom M. (1967) [1961]. Calculus. Vol. 1 (2nd ed.). John Wiley & Sons. ISBN 0-471-00005-1.
• Hardy, G. H., A Course of Pure Mathematics, Ninth edition, Cambridge University Press, 1946. (pp. 379–380).
• Rudin, Walter (1976) [1953]. Principles of mathematical analysis (3rd ed.). New York: McGraw-Hill. ISBN 0-07-054235-X. OCLC 1502474.
• Spivak, Michael (2008) [1967]. Calculus (4th ed.). Houston, TX: Publish or Perish. ISBN 978-0-914098-91-1.
• 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