Dirichlet's test

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 where M is some constant, then the series
 * $$(a_n)$$ is monotonic
 * $$\lim_{n \to \infty}a_n = 0$$
 * $$\left|\sum_{n=1}^{N}b_n\right| \leq M$$ for every positive integer N
 * $$\sum_{n=1}^{\infty} a_n b_n$$

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 $S_n = a_{n+1} B_n + \sum_{k=1}^{n} B_k (a_k - a_{k+1})$. Since $$B_n$$ is bounded by M and $$a_n \to 0$$, the first of these terms approaches zero, $$a_{n+1} B_n \to 0$$ as $$n\to\infty$$.

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

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 $$Ma_1$$ as $$n \to \infty$$. Thus, $ \sum_{k=1}^\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 $$-Ma_1$$ as $$n\to\infty$$. Thus, again, $ \sum_{k=1}^\infty M(a_k - a_{k+1})$ converges.  So, the series $ \sum_{k=1}^\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 $$b_n = (-1)^n \Longrightarrow\left|\sum_{n=1}^N b_n\right| \leq 1.$$

Another corollary is that $ \sum_{n=1}^\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.