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.