The second of the two tests mentioned in § 172 is as follows:

if $$u_{n} = \phi(n)$$ is a decreasing function of $$n$$, then the series $$\sum \phi(n)$$ is convergent or divergent according as $$\sum 2^{n}\phi(2^{n})$$ is convergent or divergent.

We can prove this by an argument which we have used already (§ 77) in the special case of the series $$\sum(1/n)$$. In the first place $\begin{gathered} \phi(3) + \phi(4) \geq 2\phi(4), \\ \phi(5) + \phi(6) + \dots + \phi(8) \geq 4\phi(8), \\ \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots \\ \phi(2^{n} + 1) + \phi(2^{n} + 2) + \dots + \phi(2^{n+1}) \geq 2^{n}\phi(2^{n+1}).\end{gathered}$ If $$\sum 2^{n}\phi(2^{n})$$ diverges then so do $$\sum 2^{n+1}\phi(2^{n+1})$$ and $$\sum 2^{n}\phi(2^{n+1})$$, and then the inequalities just obtained show that $$\sum\phi(n)$$ diverges.

On the other hand $\phi(2) + \phi(3) \leq 2\phi(2),\quad \phi(4) + \phi(5) + \dots + \phi(7) \leq 4\phi(4),$ and so on. And from this set of inequalities it follows that if $$\sum 2^{n}\phi(2^{n})$$ converges then so does $$\sum \phi(n)$$. Thus the theorem is established.

For our present purposes the field of application of this test is practically the same as that of the Integral Test. It enables us to discuss the series $$\sum n^{-s}$$ with equal ease. For $$\sum n^{-s}$$ will converge or diverge according as $$\sum 2^{n}2^{-ns}$$ converges or diverges, according as $$s > 1$$ or $$s \leq 1$$.

Example LXXII
1. Show that if $$a$$ is any positive integer greater than $$1$$ then $$\sum \phi(n)$$ is convergent or divergent according as $$\sum a^{n}\phi(a^{n})$$ is convergent or divergent. [Use the same arguments as above, taking groups of $$a$$, $$a^{2}$$, $$a^{3}$$, … terms.]

2. If $$\sum 2^{n}\phi(2^{n})$$ converges then it is obvious that $$\lim 2^{n}\phi(2^{n}) = 0$$. Hence deduce Abel’s Theorem of § 173.