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 \(\xi \to \infty\).

Example LXXI
1. Prove by an argument similar to that used above, and without integration, that \(\Phi(\xi) = \int_{1}^{\xi} \frac{dx}{x^{s}}\), where \(s < 1\), tends to infinity with \(\xi\).

2. The series \(\sum n^{-2}\), \(\sum n^{-3/2}\), \(\sum n^{-11/10}\) are convergent, and their sums are not greater than \(2\), \(3\), \(11\) respectively. The series \(\sum n^{-1/2}\), \(\sum n^{-10/11}\) are divergent.

3. The series \(\sum n^{s}/(n^{t} + a)\), where \(a > 0\), is convergent or divergent according as \(t > 1 + s\) or \(t \leq 1 + s\). [Compare with \(\sum n^{s-t}\).]

4. Discuss the convergence or divergence of the series \[\sum(a_{1}n^{s_{1}} + a_{2}n^{s_{2}} + \dots + a_{k}n^{s_{k}})/ (b_{1}n^{t_{1}} + b_{2}n^{t_{2}} + \dots + b_{l}n^{t_{l}}),\] where all the letters denote positive numbers and the \(s\)’s and \(t\)’s are rational and arranged in descending order of magnitude.

5. Prove that \[\begin{gathered} 2\sqrt{n} – 2 < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} < 2\sqrt{n} – 1, \\ \tfrac{1}{2} \pi < \frac{1}{2\sqrt{1}} + \frac{1}{3\sqrt{2}} + \frac{1}{4\sqrt{3}} + \dots < \tfrac{1}{2}(\pi + 1).\end{gathered}\]

6. If \(\phi(n) \to l > 1\) then the series \(\sum n^{-\phi(n)}\) is convergent. If \(\phi(n) \to l < 1\) then it is divergent.


$\leftarrow$ 171-174. Further tests of convergence. Abel’s Theorem. Maclaurin’s integral test Main Page 176. Cauchy’s Condensation Test $\rightarrow$
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