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 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.