The Integral Test for Positive Series Examples 2

# The Integral Test for Positive Series Examples 2

Let $f(n) = a_n$ be a continuous ultimately non-increasing positive function for the interval $[N, \infty)$ where $N \in \mathbb{N}$. Recall from The Integral Test for Positive Series page that if $\int_{N}^{\infty} f(x) \: dx$ converges then $\sum_{n=N}^{\infty} a_n$ converges. Similarly, if $\int_{N}^{\infty} f(x) \: dx$ diverges to infinity then $\sum_{n=N}^{\infty} a_n$ diverges to infinity.

We will now look at some examples of applying the integral test.

## Example 1

Determine whether the series $\sum_{n=1}^{\infty} \frac{\ln n}{n}$ converges or diverges using the integral test.

Clearly the function $f(x) = \frac{\ln x}{x}$ is continuous on $[1, \infty )$. We can easily verify that for for all $n \in \mathbb{N}$ such that $n > 1$ that $\frac{\ln n}{n} > 0$ since $\ln n > 0$ for $n > 1$ and $n > 0$ for $n ≥ 1$.

Furthermore, it's not hard to see that the sequence $\left \{ \frac{\ln n}{n} \right \}$ is ultimately decreasing since the natural logarithm function grows slower than $n$ ultimately. Therefore, we can use the integral test for this series. We evaluate the following integral:

(1)
\begin{align} \quad \int_1^{\infty} \frac{\ln x}{x} \: dx \end{align}

This integral can be evaluated with substitution. Let $u = \ln x$. Then $du = \frac{1}{x} \: dx$ and so:

(2)
\begin{align} \quad \int \frac{\ln x}{x} \: dx = \int u \: dx = \frac{u^2}{2} = \frac{\ln^2 x}{2} \end{align}

Therefore we have that:

(3)
\begin{align} \quad \int_1^{\infty} \frac{\ln x}{x} \: dx = \lim_{b \to \infty} \left [ \frac{\ln^2 x}{2} \right ]_{x=1}^{x=b} = \lim_{b \to \infty} \left ( \frac{\ln^2 b}{2} - \frac{\ln^2 1}{2} \right ) = \infty \end{align}

By the integral test, we have that since $\int_1^{\infty} \frac{\ln x}{x} \: dx$ diverges, then the series $\sum_{n=1}^{\infty} \frac{\ln n}{n}$ also diverges.

## Example 2

Determine whether the series $\sum_{n=2}^{\infty} \frac{1}{n \ln^2 n}$ converges or diverges using the integral test.

Clearly this function is continuous on $[2, \infty)$. For $n \in \mathbb{N}$ and $n ≥ 2$ this series is positive and decreasing, so we can apply the integral test.

We want to thus evaluate the following integral:

(4)
\begin{align} \quad \int_2^{\infty} \frac{1}{x \ln^2 x} \: dx \end{align}

We will need to use substitution once again to evaluate this integral. Let $u = \ln x$. Then $du = \frac{1}{x} \: dx$, and so:

(5)
\begin{align} \quad \int \frac{1}{x \ln^2 x} \: dx = \int \frac{1}{u^2} \: du = \int u^{-2} \: du = -u^{-1} = - \frac{1}{\ln x} \end{align}

Therefore we have that:

(6)
\begin{align} \quad \int_2^{\infty} \frac{1}{x \ln^2 x} \: dx = \lim_{b \to \infty} \left ( - \frac{1}{\ln x} \right )_{x=2}^{x=b} = \lim_{b \to \infty} \left ( - \frac{1}{\ln b} + \frac{1}{\ln 2} \right ) = \frac{1}{\ln 2} \end{align}

Therefore $\int_2^{\infty} \frac{1}{x \ln^2 x} \: dx$ converges and so by the integral test we have that the series $\sum_{n=2}^{\infty} \frac{1}{n \ln^2 n}$ also converges.

## Example 3

Reprove the integral test for positive series - that is, prove that if $f(n) = a_n$ where $f$ is a continuous ultimately non-increasing positive function on the interval $[N, \infty)$ where $N \in \mathbb{N}$ then if $\int_N^{\infty} f(x) \: dx$ converges then $\sum_{n=N}^{\infty} a_n$ converges, and if $\int_N^{\infty} f(x) \: dx$ diverges then $\sum_{n=N}^{\infty} a_n$ diverges.

Suppose that $\int_N^{\infty} f(x) \: dx$ converges to $L \in \mathbb{R}$.

Consider the series $\sum_{n=1}^{\infty} a_n$. Let $s_n$ be the $n^{\mathrm{th}}$ partial sum of this series - that is:

(7)
\begin{align} \quad s_n = a_1 + a_2 + ... + a_n \end{align}

Suppose that $n > N$. Then we have that:

(8)
\begin{align} \quad s_n = \underbrace{a_1 + a_2 + ... + a_N}_{=s_N} + a_{N+1} + a_{N+2} + ... + a_n \\ \quad s_n = s_N + a_{N+1} + a_{N+2} + ... + a_n \\ \quad s_n = s_N + f(N+1) + f(N+2) + ... + f(n) \end{align}

The values of $f(N+1)$, $f(N+2)$, …, $f(n)$ represent the areas of rectangles under the function $f$. Therefore:

(9)
\begin{align} \quad s_n ≤ s_N + \int_N^{\infty} f(x) \: dx \end{align}

Since $\int_N^{\infty} f(x) \: dx$ converges to $L$, we have that $0 < s_n ≤ s_N + L$. The sequence $\{ s_n \}$ is an increasing sequence and is bounded above and hence converges which implies that the series $\sum_{n=N}%{\infty} a_n$ converges.

Now suppose that $\int_N^{\infty} f(x) \: dx$ diverges. Suppose that $n > N$. This time we have that:

(10)
\begin{align} \quad s_n = \underbrace{a_1 + a_2 + ... + a_{N-1}}_{=s_{N-1}} + a_N + a_{N+1} + ... + a_n \\ \quad s_n = s_{N-1} + a_N + a_{N+1} + ... + a_n \\ \quad s_n = s_{N-1} + f(N) + f(N+1) + ... + f(n) \end{align}

The values of $f(N)$, $f(N+1)$, …, $f(n)$ overestimate the area of under $f$ and above the positive $x$ axis, and so:

(11)
\begin{align} \quad s_n ≥ s_{N-1} + \int_N^{\infty} f(x) \: dx \end{align}

Since the integral $\int_N^{\infty} f(x) \: dx$ diverges to infinity, we must have that by the comparison test that $s_n$ diverges to infinity.