The integral test

Infinite series: Convergence tests

The integral test

In the paragraph about the harmonic series we saw that $\textstyle\sum_{n=1}^\infty \frac{1}{n}=\infty$ and it was stated that $\textstyle\sum \frac{1}{n^p}$ converges if and only if $p > 1$ . In this paragraph a convergence criterium is introduced which can be used to prove this.

We will first study the series for $p = 1.1$ . This means we analyse $\textstyle\sum_{n=1}^\infty a_n$ with $\textstyle a_n = \frac{1}{n^{1.1}}$ . In the following figure the graph of $\textstyle f(x) = \frac{1}{x^{1.1}}$ is shown. Below the graph rectangles are drawn between the $x$ -axis and the graph.

The area of the first rectangle is $a_2$ , because its width is $1$ and its height equals $f(2)=\tfrac{1}{1.1^2}$ . Similarly, the area of the second rectangle is $a_3$ , the area of the third rectangle $a_4$ , etc. We know that the total area below the graph of $f$ from $x=1$ equals $\int_1^\infty f(x)\ \mathrm{d} x$ . Since the area of all rectangles together are less than the area below the graph we have $\sum_{n=2}^\infty \tfrac{1}{n^{1.1}} \leq \int_1^\infty \tfrac{1}{x^{1.1}}\ \mathrm{d}x = \int_1^\infty x^{-1.1} \mathrm{d}x= \left[ -10 x^{-0.1} \right]_1^\infty = 10\text.$ The terms of the series are positive and since the series is bounded this means it converges.

The integral of a function can also be used to show series diverge. In the following figure you can observe that $\int_1^\infty f(x)\ \mathrm{d} x \leq \sum_{n=1}^\infty a_n\text.$ The area of the first rectangle now is $a_1$ , the area of the second rectangle is $a_2$ , etc.

The next theorem generalizes this approach.

Let $(a_n)_{n=1}^\infty$ be a series such that $a_n = f(n)$ for a decreasing positive function $f\colon\mathbb R\to\mathbb R$ . Then $\sum_{n=1}^\infty a_n < \infty \iff \int_1^\infty f(x)\ \mathrm{d}x < \infty\text.$

This follows from the fact that $\sum_{k=2}^n a_k \leq \int_1^n f(x)\ \mathrm{d} x \leq \sum_{k=1}^n a_k$ for all $k$ and the squeeze lemma.

We can use this theorem to determine for which $p$ the series $\sum_{n=1}^\infty \frac{1}{n^p}$ converges.

It holds that $\sum_{n=1}^\infty \frac{1}{n^p}<\infty \iff p>1\text.$

We apply the integral test with the function $f(x) = \frac{1}{x^p}$ . Calculating the integral we obtain $\int_1^\infty f(x)\ \mathrm{d}x = \lim_{n\to\infty} \int_1^n x^{-p}\ \mathrm{d}x = \begin{cases} \lim_{n\to\infty} \left[ \log(x) \right]_{1}^n &\text{if } p = 1 \\ \lim_{n\to\infty} \left[ \frac{1}{-p+1}x^{-p+1} \right]_{1}^n &\text{if } p \neq 1\text. \end{cases}$ We see that $\lim_{n\to\infty} \left[ \log(x) \right]_{1}^n = \lim_{n\to\infty} \log(n) = \infty\text,$ so the series diverges to infinity for $p = 1$ . If $p < 1$ the limit equals $\lim_{n\to\infty} \left[ \frac{1}{-p+1}x^{-p+1} \right]_{1}^n = \lim_{n\to\infty} \left(\frac{1}{-p+1}n^{-p+1} - \frac{1}{-p+1}\right)=\infty\text,$ because the power $n$ is $-p+1 > 0$ and the constant in front of the power is positive. If $p> 1$ the power of $n$ is negative, so $n^{-p+1}\to 0$ if $n\to\infty$ . This shows the integral is finite if and only if $p > 1$ .