Dirichlet's Test for Convergence of Series of Real Numbers Examples 1

Recall from Dirichlet's Test for Convergence of Series of Real Numbers page the following test for convergence/divergence of a geometric series:

Dirichlet's Test for Convergence of Series of Real Numbers

Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be sequences of real numbers and let $(A_n)_{n=1}^{\infty}$ denote the sequence of partial sums of $\displaystyle{\sum_{n=1}^{\infty} a_n}$.

If $(A_n)_{n=1}^{\infty}$ is bounded, $(b_n)_{n=1}^{\infty}$ is a decreasing sequence of real numbers, and $\displaystyle{\lim_{n \to \infty} b_n = 0}$ then we conclude that:

• $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$ converges.

Let's now look at some examples of using Dirichlet's test.

Example 1

Show that $\displaystyle{\sum_{n=2}^{\infty} \frac{\cos(n\pi)}{\ln n}}$ converges.

Let $\displaystyle{(a_n)_{n=2}^{\infty} = (\cos(n\pi))_{n=2}^{\infty}}$ and $\displaystyle{(b_n)_{n=2}^{\infty} = \left( \frac{1}{\ln n} \right )_{n=2}^{\infty}}$.

Let $(A_n)_{n=2}^{\infty}$ be the sequence of partial sums of $(a_n)_{n=2}^{\infty}$. Notice that:

(1)

\begin{align} \quad A_2 &= \cos(2\pi) = 1 \\ \quad A_2 &= \cos(2\pi) + \cos(3\pi) = 1 - 1 = 0 \\ \quad A_3 &= \cos(\pi) + \cos(2\pi) + \cos(3\pi) = 1 - 1 + 1 = 1 \\ \quad & \vdots \end{align}

So we see that the sequence of partial sums $(A_n)_{n=2}^{\infty}$ is bounded since $\mid A_n \mid \leq 1$ for all $n \in \mathbb{N}$.

Furthermore, the sequence $\displaystyle{(b_n)_{n=2}^{\infty} = \left( \frac{1}{\ln n} \right )_{n=2}^{\infty}}$ is decreasing and $\displaystyle{\lim_{n \to \infty} \frac{1}{\ln n}} = 0$.

Therefore by Dirichlet's test for convergence, $\displaystyle{(a_n)_{n=2}^{\infty} = (\cos(n\pi))_{n=2}^{\infty}}$ converges.

Example 2

Show that $\displaystyle{\sum_{n=2}^{\infty} \frac{2^{2n}n^2}{e^nn!} \frac{1}{\ln^2 n}}$ converges.

Let $(a_n)_{n=2}^{\infty} = \left ( \frac{2^{2n}n^2}{e^nn!} \right )_{n=2}^{\infty}$ and let $\displaystyle{(b_n)_{n=2}^{\infty} = \left (\frac{1}{\ln^2 n} \right )_{n=2}^{\infty}}$.

Let $(A_n)_{n=2}^{\infty}$ be the sequence of partial sums of $(a_n)_{n=2}^{\infty}$. We claim that $(A_n)_{n=2}^{\infty}$ is bounded. One way to show this is by showing that the series $\displaystyle{\sum_{n=2}^{\infty} a_n}$ converges. Notice that:

(2)

\begin{align} \quad \rho = \lim_{n \to \infty} \left ( \frac{2^{2(n+1)}(n+1)^2}{e^{(n+1)}(n+1)!} \cdot \frac{e^nn!}{2^{2n}n^2} \right ) = \lim_{n \to \infty} \frac{4(n+1)}{en^2} = 0 \end{align}

So, by the ratio test, $\displaystyle{\sum_{n=2}^{\infty} a_n}$ converges and so $(A_n)_{n=2}^{\infty}$ converges. But every convergent sequence of real numbers is bounded. Thus $(A_n)_{n=2}^{\infty}$ is bounded.

Moreover, $(b_n)_{n=2}^{\infty}$ is a decreasing sequence such that $\displaystyle{\lim_{n \to \infty} b_n = 0}$.

So, by Dirichlet's test for convergence we have that $\displaystyle{\sum_{n=2}^{\infty} \frac{2^{2n}n^2}{e^nn!} \frac{1}{\ln^2 n}}$ converges.