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:

\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.

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