Dirichlet's Test for Convergence of Complicated Series of Real Numbs.

# 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)
\begin{align} \quad \sum_{n=1}^{\infty} \left ( \left ( \sum_{k=1}^{n} \frac{1}{k} \right ) \frac{\sin nx}{n} \right ) = \sum_{n=1}^{\infty} \left ( \left ( 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \right ) \frac{\sin nx}{n} \right ) \end{align}

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)
\begin{align} \quad a_n = \sin nx \quad \mathrm{and} \quad b_n = \frac{\left ( 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \right )}{n} \end{align}

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)
\begin{align} \quad A_n = \sum_{k=1}^{n} \sin kx = \sin x + \sin 2x + \sin 3x + ... + \sin nx \end{align}

It can be shown that:

(4)
\begin{align} \quad A_n = \frac{\cos \left ( \frac{x}{2} \right ) - \cos \left (n + \frac{1}{2} \right )x }{2 \sin \left ( \frac{x}{2} \right )} \leq \frac{1 + x}{\sin \left ( \frac{x}{2} \right )} \end{align}

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)
\begin{align} \quad b_{n+1} - b_n & = \frac{\left ( 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n+1} \right )}{n+1} - \frac{\left ( 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \right )}{n} \\ \quad b_{n+1} - b_n & = \frac{\left ( n + \frac{n}{2} + \frac{n}{3} + ... + \frac{n}{n+1} \right )}{n(n+1)} - \frac{\left ( (n+1) + \frac{(n+1)}{2} + \frac{(n+1)}{3} + ... + \frac{(n+1)}{n} \right )}{n(n+1)} \\ \quad b_{n+1} - b_n & = \frac{n-(n+1) + \frac{n-(n+1)}{2} + \frac{n-(n+1)}{3} + ... + \frac{n-(n+1)}{n} + \frac{n+1}{n}}{n(n+1)}\\ \quad b_{n+1} - b_n & = \frac{-1 -\frac{1}{2} - \frac{1}{3} - ... - \frac{1}{n} + \frac{n+1}{n}}{n(n+1)} \\ \quad b_{n+1} - b_n & \leq 0 \end{align}

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)
\begin{align} \quad \lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{1}{k} = \lim_{n \to \infty} \frac{1}{n} \left ( 1 + \frac{1}{2} + ... + \frac{1}{n} + ... \right ) = \lim_{n \to \infty} \left ( \frac{1}{n} + \frac{1}{2n} + ... + \frac{1}{n^2} + ... \right ) = 0 \end{align}

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.