By far the most important application of the Integral Test is to the series \[1^{-s} + 2^{-s} + 3^{-s} + \dots + n^{-s} + \dots,\] where \(s\) is any rational number. We have seen already (§ 77 and Ex LXVII. 14, Ex LXIX. 1) that the series is divergent when \(s = 1\).

If \(s \leq 0\) then it is obvious that the series is divergent. If \(s > 0\) then \(u_{n}\) decreases as \(n\) increases, and we can apply the test. Here \[\Phi(\xi) = \int_{1}^{\xi} \frac{dx}{x^{s}} = \frac{\xi^{1-s} – 1}{1 – s},\] unless \(s = 1\). If \(s > 1\) then \(\xi^{1-s} \to 0\) as \(\xi \to \infty\), and \[\Phi(\xi) \to \frac{1}{(s – 1)} = l,\] say. And if \(s < 1\) then \(\xi^{1-s} \to \infty\) as \(\xi \to \infty\), and so \(\Phi(\xi) \to \infty\). Thus

the series \(\sum n^{-s}\) is convergent if \(s > 1\), divergent if \(s \leq 1\), and in the first case its sum is less than \(s/(s – 1)\).

So far as divergence for \(s < 1\) is concerned, this result might have been derived at once from comparison with \(\sum (1/n)\), which we already know to be divergent.

It is however interesting to see how the Integral Test may be applied to the series \(\sum (1/n)\), when the preceding analysis fails. In this case \[\Phi(\xi) = \int_{1}^{\xi} \frac{dx}{x},\] and it is easy to see that \(\Phi(\xi) \to \infty\) as \(\xi \to \infty\). For if \(\xi > 2^{n}\) then \[\Phi(\xi) > \int_{1}^{2^{n}} \frac{dx}{x} = \int_{1}^{2} \frac{dx}{x} + \int_{2}^{4} \frac{dx}{x} + \dots + {\int_{2^{n-1}}^{2^{n}}} \frac{dx}{x}.\] But by putting \(x = 2^{r}u\) we obtain \[\int_{2^{r}}^{2^{r+1}} \frac{dx}{x} = \int_{1}^{2} \frac{du}{u},\] and so \(\Phi(\xi) > n\int_{1}^{2} \frac{du}{u}\), which shows that \(\Phi(\xi) \to \infty\) as \(\xi \to \infty\).

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