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.

Example LXXXVIII
1. The series $\sum \frac{1}{n(\log n)^{2}},\quad \sum \frac{(\log n)^{100}}{n^{101/100}},\quad \sum \frac{n^{2} – 1}{n^{2} + 1}\, \frac{1}{n(\log n)^{7/6}}$ are convergent. [The convergence of the first series is a direct consequence of the theorem of the preceding section. That of the second follows from the fact that $$(\log n)^{100}$$ is less than $$n^{\beta}$$ for sufficiently large values of $$n$$, however small $$\beta$$ may be, provided that it is positive. And so, taking $$\beta = 1/200$$, $$(\log n)^{100} n^{-101/100}$$ is less than $$n^{-201/200}$$ for sufficiently large values of $$n$$. The convergence of the third series follows from the comparison test at the end of the last section.]

2. The series $\sum \frac{1}{n(\log n)^{6/7}},\quad \sum \frac{1}{n^{100/101}(\log n)^{100}},\quad \sum \frac{n\log n}{(n\log n)^{2} + 1}$ are divergent.

3. The series $\sum \frac{(\log n)^{p}}{n^{1+s}},\quad \sum \frac{(\log n)^{p} (\log\log n)^{q}}{n^{1+s}},\quad \sum \frac{(\log\log n)^{p}}{n(\log n)^{1+s}},$ where $$s > 0$$, are convergent for all values of $$p$$ and $$q$$; similarly the series $\sum \frac{1}{n^{1-s}(\log n)^{p}},\quad \sum \frac{1}{n^{1-s}(\log n)^{p}(\log\log n)^{q}},\quad \sum \frac{1}{n(\log n)^{1-s}(\log\log n)^{p}}$ are divergent.

4. The question of the convergence or divergence of such series as $\sum \frac{1}{n\log n\log\log n},\quad \sum \frac{\log\log\log n}{n\log n\sqrtp{\log\log n}}$ cannot be settled by the theorem above, since in each case the function under the sign of summation tends to zero more rapidly than $$1/(n\log n)$$ yet less rapidly than $$n^{-1}(\log n)^{-1-\alpha}$$, where $$\alpha$$ is any positive number however small. For such series we need a still more delicate test. The reader should be able, starting from the equations \begin{aligned} D_{x}(\log_{k}x)^{1-s} &= \frac{1 – s}{x \log x \log_{2}x \dots \log_{k-1} x (\log_{k}x)^{s}},\\ D_{x}\log_{k+1}x &= \frac{1}{x \log x \log_{2}x \dots \log_{k-1}x \log_{k}x},\end{aligned} where $$\log_{2}x = \log\log x$$, $$\log_{3} x = \log\log\log x$$, …, to prove the following theorem: the series and integral $\sum_{n_{0}}^{\infty} \frac{1}{n \log n \log_{2}n \dots \log_{k-1}n (\log_{k}n)^{s}},\quad \int_{a}^{\infty} \frac{dx}{x \log x \log_{2}x \dots \log_{k-1}x (\log_{k}x)^{s}}$ are convergent if $$s > 1$$ and divergent if $$s \leq 1$$, $$n_{0}$$ and $$a$$ being any numbers sufficiently great to ensure that $$\log_{k}n$$ and $$\log_{k}x$$ are positive when $$n \geq n_{0}$$ or $$x \geq a$$. These values of $$n_{0}$$ and $$a$$ increase very rapidly as $$k$$ increases: thus $$\log x > 0$$ requires $$x > 1$$, $$\log_{2}x > 0$$ requires $$x > e$$, $${\log_{3}x} > 0$$ requires $$x > e^{e}$$, and so on; and it is easy to see that $$e^{e} > 10$$, $$e^{e^{e}} > e^{10} > 20,000$$, $$e^{e^{e^{e}}} > e^{20,000} > 10^{8000}$$.

The reader should observe the extreme rapidity with which the higher exponential functions, such as $$e^{e^{x}}$$ and $$e^{e^{e^{x}}}$$, increase with $$x$$. The same remark of course applies to such functions as $$a^{a^{x}}$$ and $$a^{a^{a^{x}}}$$, where $$a$$ has any value greater than unity. It has been computed that $$9^{9^{9}}$$ has $$369,693,100$$ figures, while $$10^{10^{10}}$$ has of course $$10,000,000,000$$. Conversely, the rate of increase of the higher logarithmic functions is extremely slow. Thus to make $$\log\log\log\log x > 1$$ we have to suppose $$x$$ a number with over $$8000$$ figures.1

5. Prove that the integral $$\int_{0}^{a} \frac{1}{x} \left\{\log \left(\frac{1}{x}\right)\right\}^{s} dx$$, where $$0 < a < 1$$, is convergent if $$s < -1$$, divergent if $$s \geq -1$$. [Consider the behaviour of $\int_{\epsilon}^{a} \frac{1}{x} \left\{\log \left(\frac{1}{x}\right)\right\}^{s} dx$ as $$\epsilon \to +0$$. This result also may be refined upon by the introduction of higher logarithmic factors.]

6. Prove that $$\int_{0}^{1} \frac{1}{x} \left\{\log \left(\frac{1}{x}\right)\right\}^{s} dx$$ has no meaning for any value of $$s$$. [The last example shows that $$s < -1$$ is a necessary condition for convergence at the lower limit: but $$\{\log(1/x)\}^{s}$$ tends to $$\infty$$ like $$(1 – x)^{s}$$, as $$x \to 1 – 0$$, if $$s$$ is negative, and so the integral diverges at the upper limit when $$s < -1$$.]

7. The necessary and sufficient conditions for the convergence of $$\int_{0}^{1} x^{a-1} \left\{\log \left(\frac{1}{x}\right)\right\}^{s} dx$$ are $$a > 0$$, $$s > -1$$.

Example LXXXIX
1. Euler’s limit. Show that $\phi(n) = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n – 1} – \log n$ tends to a limit $$\gamma$$ as $$n \to \infty$$, and that $$0 < \gamma \leq 1$$. [This follows at once from § 174. The value of $$\gamma$$ is in fact $$.577\dots$$, and $$\gamma$$ is usually called Euler’s constant.]

2. If $$a$$ and $$b$$ are positive then $\frac{1}{a} + \frac{1}{a + b} + \frac{1}{a + 2b} + \dots + \frac{1}{a + (n – 1) b} – \frac{1}{b}\log {(a + nb)}$ tends to a limit as $$n \to \infty$$.

3. If $$0 < s < 1$$ then $\phi(n) = 1 + 2^{-s} + 3^{-s} + \dots + (n – 1)^{-s} – \frac{n^{1-s}}{1 – s}$ tends to a limit as $$n \to \infty$$.

4. Show that the series $\frac{1}{1} + \frac{1}{2(1 + \frac{1}{2})} + \frac{1}{3(1 + \frac{1}{2} + \frac{1}{3})} + \dots$ is divergent. [Compare the general term of the series with $$1/(n\log n)$$.] Show also that the series derived from $$\sum n^{-s}$$, in the same way that the above series is derived from $$\sum (1/n)$$, is convergent if $$s > 1$$ and otherwise divergent.

5. Prove generally that if $$\sum u_{n}$$ is a series of positive terms, and $s_{n} = u_{1} + u_{2} + \dots + u_{n},$ then $$\sum (u_{n}/s_{n-1})$$ is convergent or divergent according as $$\sum u_{n}$$ is convergent or divergent. [If $$\sum u_{n}$$ is convergent then $$s_{n-1}$$ tends to a positive limit $$l$$, and so $$\sum (u_{n}/s_{n-1})$$ is convergent. If $$\sum u_{n}$$ is divergent then $$s_{n-1} \to \infty$$, and $u_{n}/s_{n-1} > \log\{1 + (u_{n}/s_{n-1})\} = \log (s_{n}/s_{n-1})$ (Ex. LXXXII. 1); and it is evident that $\log(s_{2}/s_{1}) + \log(s_{3}/s_{2}) + \dots + \log(s_{n}/s_{n-1}) = \log(s_{n}/s_{1})$ tends to $$\infty$$ as $$n \to \infty$$.]

6. Prove that the same result holds for the series $$\sum (u_{n}/s_{n})$$. [The proof is the same in the case of convergence. If $$\sum u_{n}$$ is divergent, and $$u_{n} < s_{n-1}$$ from a certain value of $$n$$ onwards, then $$s_{n} < 2s_{n-1}$$, and the divergence of $$\sum (u_{n}/s_{n})$$ follows from that of $$\sum (u_{n}/s_{n-1})$$. If on the other hand $$u_{n} \geq s_{n-1}$$ for an infinity of values of $$n$$, as might happen with a rapidly divergent series, then $$u_{n}/s_{n} \geq \frac{1}{2}$$ for all these values of $$n$$.]

7. Sum the series $$1 – \frac{1}{2} + \frac{1}{3} – \dots$$. [We have $1 + \frac{1}{2} + \dots + \frac{1}{2n} = \log(2n + 1) + \gamma + \epsilon_{n}, \quad 2\left(\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2n}\right) = \log(n + 1) + \gamma + \epsilon_{n}’,$ by Ex. 1, $$\gamma$$ denoting Euler’s constant, and $$\epsilon_{n}$$, $$\epsilon_{n}’$$ being numbers which tend to zero as $$n \to \infty$$. Subtracting and making $$n \to \infty$$ we see that the sum of the given series is $$\log 2$$. See also § 213.]

8. Prove that the series $\sum_{0}^{\infty} (-1)^{n}\left(1 + \frac{1}{2} + \dots + \frac{1}{n + 1} – \log n – C\right)$ oscillates finitely except when $$C = \gamma$$, when it converges.

1. See the footnote to § 202.↩︎