The Integral Test for Positive Series of Real Numbers Examples 1

Recall from The Integral Test for Positive Series of Real Numbers page the following test for convergence/divergence of a geometric series:

The Integral Test for Positive Series of Real Numbers

Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ be a positive series and let $f$ be a positive function that is decreasing and approaching $0$ on the interval $[1, \infty)$ such that $f(n) = a_n$.

a) The series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges if and only if:

• The improper integral $\displaystyle{\int_1^{\infty} f(n) \: dx}$ converges.

b) The series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges if and only if:

• The improper integral $\displaystyle{\int_1^{\infty} f(n) \: dx}$ diverges.

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

Example 1

Determine whether $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{n + 1}}$ converges or diverges.

The function $\displaystyle{f(x) = \frac{1}{x + 1}}$ is a positive, continuous, decreasing function that approaches $0$ on the interval $[1, \infty)$. So, we can apply the integral test to this series.

(1)

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

Since the improper integral $\displaystyle{\int_1^{\infty} \frac{1}{x + 1} \: dx}$ diverges, we have by the integral test that $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{n + 1}}$ diverges.

Example 2

Determine whether $\displaystyle{\sum_{n=2}^{\infty} \frac{1}{n \ln^2 n}}$ converges or diverges.

Notice that the function $\displaystyle{f(x) = \frac{1}{x \ln^2 x}}$ is a positive, continuous, decreasing function that approaches $0$ on the interval $[2, \infty)$. So, we can apply the integral test to this series. Noting that $\displaystyle{-\frac{1}{\ln b} \to 0}$ as $b \to \infty$ and we have that:

(2)

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

Since the improper integral $\displaystyle{\int_2^{\infty} \frac{1}{x \ln^2 x} \: dx}$ converges, we have by the integral test that $\displaystyle{\sum_{n=2}^{\infty} \frac{1}{n \ln^2 n}}$ converges.