Dirichlet's Test for Convergence of Complicated Series of Real Numbers
Recall from the Dirichlet's Test for Convergence of Series of Real Numbers page that if $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are two sequences of real numbers where $(A_n)_{n=1}^{\infty}$ denotes the sequence of partial sums for $(a_n)_{n=1}^{\infty}$ and if:
- $(A_n)_{n=1}^{\infty}$ is bounded.
- $(b_n)_{n=1}^{\infty}$ is a decreasing convergent sequence such that $\displaystyle{\lim_{n \to \infty} b_n = 0}$
Then we can conclude that $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$ converges.
We will now look at a complicated example of applying Dirichlet's test for convergence. Consider the following series:
(1)Suppose that we want to find all $x$ for which the series above converges. Notice that $\sin x = 0$ if $x = m\pi$ for any $m \in \mathbb{N}$, and so $\frac{\sin nx}{n} = 0$ if $x = m\pi$, and trivially, all of the terms in the series above will equal $0$ and converge to $0$. So assume that the sine function does not equal $0$ for our choice of $x$.
Using Dirichlet's test, we will show that for all $x \in \mathbb{R}$ we can establish convergence. Let:
(2)Let $(A_n)_{n=1}^{\infty}$ denote the sequence of partial sums for $(a_n)_{n=1}^{\infty}$. Then for each $n \in \mathbb{N}$ we have that:
(3)It can be shown that:
(4)Notice that the right hand side of the inequality above is fixed for every $x \in \mathbb{R}$. This shows that the sequence of partial sums $(A_n)_{n=1}^{\infty}$ is bounded.
We now establish that the sequence $(b_n)_{n=1}^{\infty}$ is decreasing. Consider the difference, $b_{n+1} - b_n$:
(5)This shows that $b_{n+1} \leq b_n$ for all $n \in \mathbb{N}$, so $(b_n)_{n=1}^{\infty}$ is a decreasing sequence. We now show that this sequence converges to $0$. Notice that:
(6)So, by Dirichlet's test we must have that the series $\displaystyle{\sum_{n=1}^{\infty} a_nb_n = \sum_{n=1}^{\infty} \left ( \left ( 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \right ) \frac{\sin nx}{n} \right )}$ converges.