Dirichlet's test

Statement

The test states that if { a n } is a sequence of real numbers and { b n } a sequence of complex numbers satisfying

where M is some constant, then the series

converges.

Proof

Let S n = ∑ k = 1 n a k b k and B n = ∑ k = 1 n b k .

From summation by parts, we have that S n = a n + 1 B n + ∑ k = 1 n B k ( a k − a k + 1 ) .

Since B n is bounded by M and a n → 0 , the first of these terms approaches zero, a n + 1 B n → 0 as n→∞.

On the other hand, since the sequence a n is decreasing, a k − a k + 1 is positive for all k, so | B k ( a k − a k + 1 ) | ≤ M ( a k − a k + 1 ) . That is, the magnitude of the partial sum of B_n, times a factor, is less than the upper bound of the partial sum B_n (a value M) times that same factor.

But ∑ k = 1 n M ( a k − a k + 1 ) = M ∑ k = 1 n ( a k − a k + 1 ) , which is a telescoping series that equals M ( a 1 − a n + 1 ) and therefore approaches M a 1 as n→∞. Thus, ∑ k = 1 ∞ M ( a k − a k + 1 ) converges.

In turn, ∑ k = 0 ∞ | B k ( a k − a k + 1 ) | converges as well by the Direct comparison test. The series ∑ k = 1 ∞ B k ( a k − a k + 1 ) converges, as well, 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

Another corollary is that ∑ n = 1 ∞ a n sin ⁡ n converges whenever { a n } is a decreasing sequence that tends to zero.

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