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.↩︎

$\leftarrow$ 210. Common logarithms Main Page 212. Series connected with the exponential and logarithmic functions. Expansion of \(e^{x}\) by Taylor’s Theorem $\rightarrow$
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