175. The series $\sum n^{-s}$

By far the most important application of the Integral Test is to the series $$1^{-s}+2^{-s}+3^{-s}+\cdots +n^{-s}+\dots ,$$ 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 $s\le 0$ 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 $$\mathrm{\Phi }(\xi )={\int }_1^{\xi }\frac{dx}{x^s}=\frac{{\xi }^{1-s}–1}{1–s},$$ unless $s=1$. If $s>1$ then ${\xi }^{1-s}\to 0$ as $\xi \to \mathrm{\infty }$, and $$\mathrm{\Phi }(\xi )\to \frac{1}{(s–1)}=l,$$ say. And if $s<1$ then ${\xi }^{1-s}\to \mathrm{\infty }$ as $\xi \to \mathrm{\infty }$, and so $\mathrm{\Phi }(\xi )\to \mathrm{\infty }$. Thus

the series $\sum n^{-s}$ is convergent if $s>1$, divergent if $s\le 1$, 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 $$\mathrm{\Phi }(\xi )={\int }_1^{\xi }\frac{dx}{x},$$ and it is easy to see that $\mathrm{\Phi }(\xi )\to \mathrm{\infty }$ as $\xi \to \mathrm{\infty }$. For if $\xi >2^n$ then $$\mathrm{\Phi }(\xi )>{\int }_1^{2^n}\frac{dx}{x}={\int }_1^2\frac{dx}{x}+{\int }_2^4\frac{dx}{x}+\cdots +{\int }_{2^{n-1}}^{2^n}\frac{dx}{x}.$$ 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 $\mathrm{\Phi }(\xi )>n{\int }_1^2\frac{du}{u}$, which shows that $\mathrm{\Phi }(\xi )\to \mathrm{\infty }$ as $\xi \to \mathrm{\infty }$.