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.

Het integraalkenmerkLet \((a_n)_{n=1}^\infty\) 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 \(p > 1\).