The Integral Test for Positive Series Examples 1

# The Integral Test for Positive Series Examples 1

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

Using the integral test, determine whether the series $\sum_{n=1}^{\infty} \frac{1}{1 + 10n}$ is convergent or divergent.

The answer to this question should be fairly obvious using the divergence theorem (since $\lim_{n \to \infty} a_n = 1$), but let's show it with the integral test. Let $f(x) = \frac{1}{1 + 10x}$. This function is positive for $x ≥ 1$, non-increasing, and continuous. Let $u = 1 + 10x$ and so $du = 10 \: dx$ which implies $\frac{du}{10} = dx$, and so:

(1)
\begin{align} \quad \int \frac{1}{1 + 10x} \: dx = \frac{1}{10} \int \frac{1}{u} \: du = \frac{1}{10} \ln u + C = \frac{1}{10} \ln (1 + 10x) + C \end{align}

Therefore it follows that:

(2)
\begin{align} \quad \int_{1}^{\infty} \frac{1}{1 + 10x} \: dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{1 + 10x} = \lim_{b \to \infty} \left ( \frac{1}{10} \ln (1 + 10x) + C \right)_{1}^{b} = \lim_{b \to \infty} \left (\frac{1}{10} \ln (1 + 10b) - \frac{1}{10}\ln 11 \right) = \infty \end{align}

Therefore the series $\sum_{n=1}^{\infty} \frac{1}{1 + 10n}$ is divergent.

## Example 2

Using the integral test, determine whether the series $\sum_{n=1}^{\infty} \frac{n}{1 + n^2}$ is convergent or divergent.

First let $f(x) = \frac{x}{1 + x^2 }$. This function is positive for $x ≥ 1$, non-increasing, and continuous. We now evaluate the following integral by letting $u = 1 + x^2$ so that $du = 2x \: dx$ which implies that $\frac{du}{2} = x \: dx$, and so:

(3)
\begin{align} \quad \int \frac{x}{1 + x^2} = \frac{1}{2} \int \frac{1}{u} \: du = \frac{1}{2} \ln u + C = \frac{1}{2} \ln (1 + x^2) + C \end{align}

Therefore it follows that:

(4)
\begin{align} \quad \int_{1}^{\infty} \frac{x}{1 + x^2} = \lim_{b \to \infty} \int_{1}^{b} \frac{x}{1 + x^2} = \lim_{b \to \infty} \left ( \frac{1}{2} \ln (1 + x^2) \right)_{1}^{b} = \lim_{b \to \infty} \left ( \frac{1}{2} \ln (1 + b^2) - \frac{1}{2} \ln (2) \right) = \infty \end{align}

## Example 3

Using the integral test, determine whether the series $\sum_{n=1}^{\infty} ne^{-n}$ is convergent or divergent.

First let $f(x) = xe^{-x}$. This function is positive for $x ≥ 1$, non-increasing, and continuous. We now evaluate the following integral with integration by parts. Let $u = x$ and let $dv = e^{-x} \: dx$ and so $du = dx$ and $v = -e^{-x}$

(5)
\begin{align} \int xe^{-x} \: dx = -xe^{-x} + \int e^{-x} \: dx = -xe^{-x} -e^{-x} + C \end{align}

And therefore:

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

We now want to evaluate $\lim_{x \to \infty} \frac{x + 1}{e^x}$ using L'Hospital's rule, and so:

(7)
\begin{align} \lim_{x \to \infty} \frac{x + 1}{e^x} = \lim_{x \to \infty} \frac{1}{e^x} = 0 \end{align}

Therefore $- \lim_{b \to \infty} \left ( \frac{b + 1}{e^b} - \frac{2}{e} \right ) = \frac{2}{e}$ and therefore $\sum_{n=1}^{\infty} ne^{-n}$ is convergent.