Dirichlet's test

Short description: 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.

Statement

The test states that if $(a_n)$ is a sequence of real numbers and $(b_n)$ a sequence of complex numbers satisfying

• $(a_n)$ is monotonic
• $\underset{n\to \mathrm{\infty }}{\lim }{a}_n=0$
• $\left|\sum _{n=1}^N{b}_n\right|\le M$ for every positive integer N

where M is some constant, then the series

$$\sum _{n=1}^{\mathrm{\infty }}{a}_n{b}_n$$

converges.

Proof

Let $S_n={\sum }_{k=1}^n{a}_k{b}_k$ and $B_n={\sum }_{k=1}^n{b}_k$.

From summation by parts, we have that $S_n=a_n{B}_n+{\sum }_{k=1}^{n-1}{B}_k(a_k-a_{k+1})$. Since $B_n$ is bounded by M and $a_n\to 0$, the first of these terms approaches zero, $a_n{B}_n\to 0$ as $n\to \mathrm{\infty }$.

We have, for each k, $|B_k(a_k-a_{k+1})|\le M|a_k-a_{k+1}|$.

Since $(a_n)$ is monotone, it is either decreasing or increasing:

• If $(a_n)$ is decreasing, $$\sum _{k=1}^{n}M|a_k-a_{k+1}|=\sum _{k=1}^{n}M(a_k-a_{k+1})=M\sum _{k=1}^n(a_k-a_{k+1}),$$ which is a telescoping sum that equals $M(a_1-a_{n+1})$ and therefore approaches $M{a}_1$ as $n\to \mathrm{\infty }$. Thus, ${\sum }_{k=1}^{\mathrm{\infty }}M(a_k-a_{k+1})$ converges.
• If $(a_n)$ is increasing, $$\sum _{k=1}^{n}M|a_k-a_{k+1}|=-\sum _{k=1}^{n}M(a_k-a_{k+1})=-M\sum _{k=1}^n(a_k-a_{k+1}),$$ which is again a telescoping sum that equals $-M(a_1-a_{n+1})$ and therefore approaches $-M{a}_1$ as $n\to \mathrm{\infty }$. Thus, again, ${\sum }_{k=1}^{\mathrm{\infty }}M(a_k-a_{k+1})$ converges.

So, the series ${\sum }_{k=1}^{\mathrm{\infty }}{B}_k(a_k-a_{k+1})$ converges, by the absolute convergence test. Hence $S_n$ converges.

Applications

A particular case of Dirichlet's test is the more commonly used alternating series test for the case $$b_n=(-1{)}^n⟹\left|\sum _{n=1}^N{b}_n\right|\le 1.$$

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

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.

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