175. The series (sum n^{-s})

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

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176. Cauchy’s Condensation Test $\rightarrow$