211. Logarithmic tests of convergence for series and integrals

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

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

4. The question of the convergence or divergence of such series as $$\sum \frac{1}{n\log n\log \log n},\sum \frac{\log \log \log n}{n\log n\text{sqrtp}\log \log n}$$ 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 $$\begin{array}{rl}{D}_x({\log }_{k}x{)}^{1-s}& =\frac{1–s}{x\log x{\log }_{2}x\dots {\log }_{k-1}x({\log }_{k}x{)}^s},\\ D_x{\log }_{k+1}x& =\frac{1}{x\log x{\log }_{2}x\dots {\log }_{k-1}x{\log }_{k}x},\end{array}$$ where ${\log }_{2}x=\log \log x$, ${\log }_{3}x=\log \log \log x$, …, to prove the following theorem: the series and integral $$\sum _{n_0}^{\mathrm{\infty }}\frac{1}{n\log n{\log }_{2}n\dots {\log }_{k-1}n({\log }_{k}n{)}^s},{\int }_a^{\mathrm{\infty }}\frac{dx}{x\log x{\log }_{2}x\dots {\log }_{k-1}x({\log }_{k}x{)}^s}$$ are convergent if $s>1$ and divergent if $s\le 1$, $n_0$ and $a$ being any numbers sufficiently great to ensure that ${\log }_{k}n$ and ${\log }_{k}x$ are positive when $n\ge n_0$ or $x\ge a$. 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 ${\int }_0^a\frac{1}{x}{\{\log \left(\frac{1}{x}\right)\}}^{s}dx$, where $0<a<1$, is convergent if $s<-1$, divergent if $s\ge -1$. [Consider the behaviour of $${\int }_{ϵ}^a\frac{1}{x}{\{\log \left(\frac{1}{x}\right)\}}^{s}dx$$ as $ϵ\to +0$. This result also may be refined upon by the introduction of higher logarithmic factors.]

6. Prove that ${\int }_0^1\frac{1}{x}{\{\log \left(\frac{1}{x}\right)\}}^{s}dx$ 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 $\mathrm{\infty }$ 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 ${\int }_0^1{x}^{a-1}{\{\log \left(\frac{1}{x}\right)\}}^{s}dx$ are $a>0$, $s>-1$.

1. Euler’s limit. Show that $$\varphi (n)=1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n–1}–\log n$$ tends to a limit $\gamma$ as $n\to \mathrm{\infty }$, and that $0<\gamma \le 1$. [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 $$\frac{1}{a}+\frac{1}{a+b}+\frac{1}{a+2b}+\cdots +\frac{1}{a+(n–1)b}–\frac{1}{b}\log (a+nb)$$ tends to a limit as $n\to \mathrm{\infty }$.

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

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 $$s_n=u_1+u_2+\cdots +u_n,$$ 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 $s_{n-1}\to \mathrm{\infty }$, 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 $$\log (s_2/s_1)+\log (s_3/s_2)+\cdots +\log (s_n/s_{n-1})=\log (s_n/s_1)$$ tends to $\mathrm{\infty }$ as $n\to \mathrm{\infty }$.]

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 $u_n\ge s_{n-1}$ for an infinity of values of $n$, as might happen with a rapidly divergent series, then $u_n/s_n\ge \frac{1}{2}$ for all these values of $n$.]

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

8. Prove that the series $$\sum _0^{\mathrm{\infty }}(-1{)}^n(1+\frac{1}{2}+\cdots +\frac{1}{n+1}–\log n–C)$$ oscillates finitely except when $C=\gamma$, when it converges.