We showed in Ch.VIII (§ 175 *et seq.*) that \[\sum_{1}^{\infty} \frac{1}{n^{s}},\quad \int_{a}^{\infty} \frac{dx}{x^{s}}\qquad (a > 0)\] are convergent if \(s > 1\) and divergent if \(s \leq 1\). Thus \(\sum (1/n)\) is divergent, but \(\sum n^{-1-\alpha}\) is convergent for all positive values of \(\alpha\).

We saw however in § 200 that with the aid of logarithms we can construct functions which tend to zero, as \(n \to \infty\), more rapidly than \(1/n\), yet less rapidly than \(n^{-1-\alpha}\), however small \(\alpha\) may be, provided of course that it is positive. For example \(1/(n\log n)\) is such a function, and the question as to whether the series \[\sum \frac{1}{n\log n}\] is convergent or divergent cannot be settled by comparison with any series of the type \(\sum n^{-s}\).

The same is true of such series as \[\sum \frac{1}{n(\log n)^{2}},\quad \sum \frac{\log\log n}{n\sqrt{\log n}}.\] It is a question of some interest to find tests which shall enable us to decide whether series such as these are convergent or divergent; and such tests are easily deduced from the Integral Test of § 174.

For since \[D_{x}(\log x)^{1-s} = \frac{1 – s}{x(\log x)^{s}},\quad D_{x}\log\log x = \frac{1}{x\log x},\] we have \[\int_{a}^{\xi} \frac{dx}{x(\log x)^{s}} = \frac{(\log\xi)^{1-s} – (\log a)^{1-s}}{1 – s},\quad \int_{{a}}^{\xi} \frac{dx}{x\log x} = \log\log \xi – \log\log a,\] if \(a > 1\). The first integral tends to the limit \(-(\log a)^{1-s}/(1 – s)\) as \(\xi \to \infty\), if \(s > 1\), and to \(\infty\) if \(s < 1\). The second integral tends to \(\infty\). Hence

*the series and integral \[\sum_{n_{0}}^{\infty} \frac{1}{n(\log n)^{s}},\quad \int_{a}^{\infty} \frac{dx}{x(\log x)^{s}},\] where \(n_{0}\) and \(a\) are greater than unity, are convergent if \(s > 1\), divergent if \(s \leq 1\).*

It follows, of course, that \(\sum \phi(n)\) is convergent if \(\phi(n)\) is positive and less than \(K/\{n(\log n)^{s}\}\), where \(s > 1\), for all values of \(n\) greater than some definite value, and divergent if \(\phi(n)\) is positive and greater than \(K/(n\log n)\) for all values of \(n\) greater than some definite value. And there is a corresponding theorem for integrals which we may leave to the reader.

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