Dirichlet's Test for Convergence of Series of Real Numbers

# Dirichlet's Test for Convergence of Series of Real Numbers

Recall from The Partial Summation Formula for Series of Real Numbers page that if $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are both sequences of real numbers then:

(1)\begin{align} \quad \sum_{k=1}^{n} a_kb_k = A_nb_{n+1} - \sum_{k=1}^{n} A_k(b_{k+1} - b_k) \end{align}

Furthermore we noted that the series $\displaystyle{\sum_{k=1}^{\infty} a_kb_k}$ converges if $(A_nb_{n+1})_{n=1}^{\infty}$ converges and $\displaystyle{\sum_{k=1}^{\infty} A_k(b_{k+1} - b_k)}$ converges.

We will now look at a very important test for convergence known as Dirichlet's test which relies on the theorem above.

Theorem 1 (Dirichlet's Test): Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be two sequences of real numbers. If the sequence of partial sums of $(a_n)_{n=1}^{\infty}$, denote it $(A_n)_{n=1}^{\infty}$, is bounded and $(b_n)_{n=1}^{\infty}$ is a decreasing sequence such that $\displaystyle{\lim_{n \to \infty} b_n = 0}$ then $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$ converges. |

**Proof:**Let $(A_n)_{n=1}^{\infty}$ be a bounded sequence. Then there exists an $M \in \mathbb{R}$, $M > 0$ such that for all $n \in \mathbb{N}$ we have that:

\begin{align} \quad \mid A_n \mid \leq M \end{align}

- Therefore we see that:

\begin{align} \quad \lim_{n \to \infty} -M b_{n+1} \leq \lim_{n \to \infty} - \mid A_n \mid b_{n+1} \leq \lim_{n \to \infty} A_nb_{n+1} \leq \lim_{n \to \infty} \mid A_n \mid b_{n+1} \leq \lim_{n \to \infty} M b_{n+1} \end{align}

- Since $\displaystyle{\lim_{n \to \infty} b_n = 0}$, we have by the Squeeze theorem that $\displaystyle{\lim_{n \to \infty} A_nb_{n+1} = 0}$.

- Furthermore we have that:

\begin{align} \quad \sum_{n=1}^{\infty} \mid A_n (b_{n+1} - b_n) \mid = \sum_{n=1}^{\infty} \mid A_n \mid \mid b_{n+1} - b_n \mid \leq \sum_{n=1}^{\infty} M \mid b_{n+1} - b_n \mid = M \sum_{n=1}^{\infty} (b_{n+1} - b_n) \end{align}

- Let $\displaystyle{s_n = \sum_{k=1}^{n} (b_{k+1} - b_k) = b_{n+1} - b_1}$. So $\displaystyle{\lim_{n \to \infty} s_n = \lim_{n \to \infty} b_{n+1} - b_1 = - b_1}$. Therefore the series $\displaystyle{\sum_{n=1}^{\infty} (b_{n+1} - b_n)}$ converges and by the comparison test we have that $\displaystyle{\sum_{n=1}^{\infty} \mid A_n (b_{n+1} - b_n) \mid}$ converges. This implies that $\displaystyle{\sum_{n=1}^{\infty} A_n(b_{n+1} - b_n)}$ converges absolutely and is hence convergent.

- Since the terms on the righthand side of the partial summation formula for the sequences $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ converge we see that also $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$ converges. $\blacksquare$

It is very important to note that the alternating series test is a nice corollary of the result above.