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In mathematics, Dirichlet's test is a method of testing for the convergence of a series that is especially useful for proving conditional convergence. 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]

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