1. The function \(\phi(n)\) takes the values \(1\), \(0\), \(0\), \(0\), \(1\), \(0\), \(0\), \(0\), \(1\), … when \(n = 0\), \(1\), \(2\), …. Express \(\phi(n)\) in terms of \(n\) by a formula which does not involve trigonometrical functions. [\(\phi(n) = \frac{1}{4}\{1 + (-1)^{n} + i^{n} + (-i)^{n}\}\).]


2. If \(\phi(n)\) steadily increases, and \(\psi(n)\) steadily decreases, as \(n\) tends to \(\infty\), and if \(\psi(n) > \phi(n)\) for all values of \(n\), then both \(\phi(n)\) and \(\psi(n)\) tend to limits, and \(\lim\phi(n) \leq \lim\psi(n)\). [This is an immediate corollary from § 69.]


3. Prove that, if \[\phi(n) = \left(1 + \frac{1}{n}\right)^{n},\quad \psi(n) = \left(1 – \frac{1}{n}\right)^{-n},\] then \(\phi(n + 1) > \phi(n)\) and \(\psi(n + 1) < \psi(n)\). [The first result has already been proved in § 73.]


4. Prove also that \(\psi(n) > \phi(n)\) for all values of \(n\): and deduce (by means of the preceding examples) that both \(\phi(n)\) and \(\psi(n)\) tend to limits as \(n\) tends to \(\infty\).1


5. The arithmetic mean of the products of all distinct pairs of positive integers whose sum is \(n\) is denoted by \(S_{n}\). Show that \(\lim(S_{n}/n^{2}) = 1/6\).


6. Prove that if \(x_{1} = \frac{1}{2}\{x + (A/x)\}\), \(x_{2} = \frac{1}{2}\{x_{1} + (A/x_{1})\}\), and so on, \(x\) and \(A\) being positive, then \(\lim x_{n} = \sqrt{A}\).

[Prove first that \(\dfrac{x_{n} – \sqrt{A}}{x_{n} + \sqrt{A}} = \biggl(\dfrac{x – \sqrt{A}}{x + \sqrt{A}}\biggr)^{2^{n}}\).]


7. If \(\phi(n)\) is a positive integer for all values of \(n\), and tends to \(\infty\) with \(n\), then \(x^{\phi(n)}\) tends to \(0\) if \(0 < x < 1\) and to \(+\infty\) if \(x > 1\). Discuss the behaviour of \(x^{\phi(n)}\), as \(n \to \infty\), for other values of \(x\).


8. If \(a_{n}\) increases or decreases steadily as \(n\) increases, then the same is true of \((a_{1} + a_{2} + \dots + a_{n})/n\).


9. If \(x_{n+1} = \sqrt{k + x_{n}}\), and \(k\) and \(x_{1}\) are positive, then the sequence \(x_{1}\), \(x_{2}\), \(x_{3}\), … is an increasing or decreasing sequence according as \(x_{1}\) is less than or greater than \(\alpha\), the positive root of the equation \(x^{2} = x + k\); and in either case \(x_{n} \to \alpha\) as \(n \to \infty\).


10. If \(x_{n+1} = k/(1 + x_{n})\), and \(k\) and \(x_{1}\) are positive, then the sequences \(x_{1}\), \(x_{3}\), \(x_{5}\), … and \(x_{2}\), \(x_{4}\), \(x_{6}\), … are one an increasing and the other a decreasing sequence, and each sequence tends to the limit \(\alpha\), the positive root of the equation \(x^{2} + x = k\).


11. The function \(f(x)\) is increasing and continuous (see Ch. V) for all values of \(x\), and a sequence \(x_{1}\), \(x_{2}\), \(x_{3}\), … is defined by the equation \(x_{n+1} = f(x_{n})\). Discuss on general graphical grounds the question as to whether \(x_{n}\) tends to a root of the equation \(x = f(x)\). Consider in particular the case in which this equation has only one root, distinguishing the cases in which the curve \(y = f(x)\) crosses the line \(y = x\) from above to below and from below to above.


12. If \(x_{1}\), \(x_{2}\) are positive and \(x_{n+1} = \frac{1}{2} (x_{n} + x_{n-1})\), then the sequences \(x_{1}\), \(x_{3}\), \(x_{5}\), … and \(x_{2}\), \(x_{4}\), \(x_{6}\), … are one a decreasing and the other an increasing sequence, and they have the common limit \(\frac{1}{3}(x_{1} + 2x_{2})\).


13. Draw a graph of the function \(y\) defined by the equation \[y = \lim_{n \to \infty} \frac{x^{2n} \sin\frac{1}{2}\pi x + x^{2}}{x^{2n} + 1}.\]


14. The function \[y = \lim_{n \to \infty} \frac{1}{1 + n\sin^{2} \pi x}\] is equal to \(0\) except when \(x\) is an integer, and then equal to \(1\). The function \[y = \lim_{n \to \infty} \frac{\psi(x) + n\phi(x) \sin^{2}\pi x}{1 + n\sin^{2}\pi x}\] is equal to \(\phi(x)\) unless \(x\) is an integer, and then equal to \(\psi(x)\).


15. Show that the graph of the function \[y = \lim_{n \to \infty} \frac{x^{n}\phi(x) + x^{-n}\psi(x)}{x^{n} + x^{-n}}\] is composed of parts of the graphs of \(\phi(x)\) and \(\psi(x)\), together with (as a rule) two isolated points. Is \(y\) defined when (a) \(x = 1\), (b) \(x = -1\), (c) \(x = 0\)?


16. Prove that the function \(y\) which is equal to \(0\) when \(x\) is rational, and to \(1\) when \(x\) is irrational, may be represented in the form \[y = \lim_{m \to \infty} \operatorname{sgn}\{\sin^{2}(m!\, \pi x)\},\] where \[\operatorname{sgn} x = \lim_{n \to \infty} (2/\pi)\arctan(nx),\] as in Ex. XXXI. 14. [If \(x\) is rational then \(\sin^{2}(m!\, \pi x)\), and therefore \(\operatorname{sgn}\{\sin^{2}(m!\, \pi x)\}\), is equal to zero from a certain value of \(m\) onwards: if \(x\) is irrational then \(\sin^{2}(m!\, \pi x)\) is always positive, and so \(\operatorname{sgn}\{\sin^{2}(m!\, \pi x)\}\) is always equal to \(1\).]

Prove that \(y\) may also be represented in the form \[1 – \lim_{m \to\infty} [\lim_{n \to\infty}\{\cos(m!\, \pi x)\}^{2n}].\]


17. Sum the series \[\sum_{1}^{\infty} \frac{1}{\nu(\nu + 1)},\quad \sum_{1}^{\infty} \frac{1}{\nu(\nu + 1)\dots(\nu + k)}.\]

[Since \[\frac{1}{\nu(\nu + 1)\dots(\nu + k)} = \frac{1}{k} \biggl\{\frac{1}{\nu(\nu + 1)\dots(\nu + k – 1)} – \frac{1}{(\nu + 1)(\nu + 2)\dots(\nu + k)}\biggr\},\] we have \[\sum_{1}^{n} \frac{1}{\nu(\nu + 1)\dots(\nu + k)} = \frac{1}{k}\left\{\frac{1}{1\cdot2\dots k} – \frac{1}{(n + 1)(n + 2)\dots (n + k)}\right\}\] and so \[\sum_{1}^{\infty} \frac{1}{\nu(\nu + 1)\dots (\nu + k)} = \frac{1}{k(k!)}.]\]


18. If \(|z| < |\alpha|\), then \[\begin{aligned} \frac{L}{z – \alpha} = -&\frac{L}{\alpha}\left(1 + \frac{z}{\alpha} + \frac{z^{2}}{\alpha^{2}} + \dots\right);\end{aligned}\] and if $|z|>|\alpha|$, then \[\begin{aligned} \frac{L}{z – \alpha} = &\frac{L}{z}\left(1 + \frac{\alpha}{z} + \frac{\alpha^{2}}{z^{2}} + \dots\right).\end{aligned}\]


19. Expansion of \((Az + B)/(az^{2} + 2bz + c)\) in powers of \(z\). Let \(\alpha\), \(\beta\) be the roots of \(az^{2} + 2bz + c = 0\), so that \(az^{2} + 2bz + c = a(z – \alpha)(z – \beta)\). We shall suppose that \(A\), \(B\), \(a\), \(b\), \(c\) are all real, and \(\alpha\) and \(\beta\) unequal. It is then easy to verify that \[\frac{Az + B}{az^{2} + 2bz + c} = \frac{1}{a(\alpha – \beta)} \left(\frac{A\alpha + B}{z – \alpha} – \frac{A\beta + B}{z – \beta}\right).\] There are two cases, according as \(b^{2} > ac\) or \(b^{2} < ac\).

(1) If \(b^{2} > ac\) then the roots \(\alpha\), \(\beta\) are real and distinct. If \(|z|\) is less than either \(|\alpha|\) or \(|\beta|\) we can expand \(1/(z – \alpha)\) and \(1/(z – \beta)\) in ascending powers of \(z\) (Ex. 18). If \(|z|\) is greater than either \(|\alpha|\) or \(|\beta|\) we must expand in descending powers of \(z\); while if \(|z|\) lies between \(|\alpha|\) and \(|\beta|\) one fraction must be expanded in ascending and one in descending powers of \(z\). The reader should write down the actual results. If \(|z|\) is equal to \(|\alpha|\) or \(|\beta|\) then no such expansion is possible.

(2) If \(b^{2} < ac\) then the roots are conjugate complex numbers (Ch. III § 43), and we can write \[\alpha = \rho\operatorname{Cis}\phi, \quad \beta = \rho\operatorname{Cis}(-\phi),\] where \(\rho^{2} = \alpha\beta = c/a\), \(\rho\cos\phi = \frac{1}{2}(\alpha + \beta) = – b/a\), so that \(\cos\phi = -\sqrt{b^{2}/ac}\), \(\sin\phi = \sqrt{1 – (b^{2}/ac)}\).

If \(|z| < \rho\) then each fraction may be expanded in ascending powers of \(z\). The coefficient of \(z^{n}\) will be found to be \[\frac{A\rho\sin n\phi + B\sin\{(n + 1)\phi\}}{a\rho^{n+1} \sin\phi}.\] If \(|z| > \rho\) we obtain a similar expansion in descending powers, while if \(|z| = \rho\) no such expansion is possible.


20. Show that if \(|z| < 1\) then \[1 + 2z + 3z^{2} + \dots + (n + 1)z^{n} + \dots = 1/(1 – z)^{2}.\]

[The sum to \(n\) terms is \(\dfrac{1 – z^{n}}{(1 – z)^{2}} – \dfrac{nz^{n}}{1 – z}\).]


21. Expand \(L/(z – \alpha)^{2}\) in powers of \(z\), ascending or descending according as \(|z| < |\alpha|\) or \(|z| > |\alpha|\).


22. Show that if \(b^{2} = ac\) and \(|az| < |b|\) then \[\frac{Az + B}{az^{2} + 2bz + c} = \sum_{0}^{\infty} p_{n}z^{n},\] where \(p_{n} = \{(-a)^{n}/b^{n+2}\} \{(n + 1)aB – nbA\}\); and find the corresponding expansion, in descending powers of \(z\), which holds when \(|az| > |b|\).


23. Verify the result of Ex. 19 in the case of the fraction \(1/(1 + z^{2})\). [We have \(1/(1 + z^{2}) = \sum z^{n} \sin\{\frac{1}{2}(n + 1)\pi\} = 1 – z^{2} + z^{4} – \dots\).]


24. Prove that if \(|z| < 1\) then \[\frac{1}{1 + z + z^{2}} = \frac{2}{\sqrt{3}} \sum_{0}^{\infty} z^{n} \sin\{\tfrac{2}{3}(n + 1)\pi\}.\]


25. Expand \((1 + z)/(1 + z^{2})\), \((1 + z^{2})/(1 + z^{3})\) and \((1 + z + z^{2})/(1 + z^{4})\) in ascending powers of \(z\). For what values of \(z\) do your results hold?


26. If \(a/(a + bz + cz^{2}) = 1 + p_{1}z + p_{2}z^{2} + \dots\) then \[1 + p_{1}^{2}z + p_{2}^{2}z^{2} + \dots = \frac{a + cz}{a – cz}\, \frac{a^{2}}{a^{2} – (b^{2} – 2ac)z + c^{2}z^{2}}.\]


27. If \(\lim\limits_{n \to \infty} s_{n} = l\) then \[\lim_{n \to \infty} \frac{s_{1} + s_{2} + \dots + s_{n}}{n} = l.\]

[Let \(s_{n} = l + t_{n}\). Then we have to prove that \((t_{1} + t_{2} + \dots + t_{n})/n\) tends to zero if \(t_{n}\) does so.

We divide the numbers \(t_{1}\), \(t_{2}\), … \(t_{n}\) into two sets \(t_{1}\), \(t_{2}\), …, \(t_{p}\) and \(t_{p+1}\), \(t_{p+2}\), …, \(t_{n}\). Here we suppose that \(p\) is a function of \(n\) which tends to \(\infty\) as \(n \to \infty\), but more slowly than \(n\), so that \(p \to \infty\) and \(p/n \to 0\): we might suppose \(p\) to be the integral part of \(\sqrt{n}\).

Let \({\epsilon}\) be any positive number. However small \(\epsilon\) may be, we can choose \(n_{0}\) so that \(t_{p+1}\), \(t_{p+2}\), …, \(t_{n}\) are all numerically less than \(\frac{1}{2}\epsilon\) when \(n \geq n_{0}\), and so \[|(t_{p+1} + t_{p+2} + \dots + t_{n})/n| < \tfrac{1}{2}\epsilon(n – p)/n < \tfrac{1}{2} \epsilon.\] But, if \(A\) is the greatest of the moduli of all the numbers \(t_{1}\), \(t_{2}\), …, we have \[|(t_{1} + t_{2} + \dots + t_{p})/n| < pA/n,\] and this also will be less than \(\frac{1}{2}\epsilon\) when \(n \geq n_{0}\), if \(n_{0}\) is large enough, since \(p/n \to 0\) as \(n \to \infty\). Thus \[|(t_{1} + t_{2} + \dots + t_{n})/n| \leq |(t_{1} + t_{2} + \dots + t_{p})/n| + |(t_{p+1} + \dots + t_{n})/n| < \epsilon\] when \(n \geq n_{0}\); which proves the theorem.

The reader, if he desires to become expert in dealing with questions about limits, should study the argument above with great care. It is very often necessary, in proving the limit of some given expression to be zero, to split it into two parts which have to be proved to have the limit zero in slightly different ways. When this is the case the proof is never very easy.

The point of the proof is this: we have to prove that \((t_{1} + t_{2} + \dots + t_{n})/n\) is small when \(n\) is large, the \(t\)’s being small when their suffixes are large. We split up the terms in the bracket into two groups. The terms in the first group are not all small, but their number is small compared with \(n\). The number in the second group is not small compared with \(n\), but the terms are all small, and their number at any rate less than \(n\), so that their sum is small compared with \(n\). Hence each of the parts into which \((t_{1} + t_{2} + \dots + t_{n})/n\) has been divided is small when \(n\) is large.]


28. If \(\phi(n) – \phi(n – 1)\to l\) as \(n \to \infty\), then \(\phi(n)/n \to l\).

[If \(\phi(n) = s_{1} + s_{2} + \dots + s_{n}\) then \(\phi(n) – \phi(n – 1) = s_{n}\), and the theorem reduces to that proved in the last example.]


29. If \(s_{n} = \frac{1}{2}\{1 – (-1)^{n}\}\), so that \(s_{n}\) is equal to \(1\) or \(0\) according as \(n\) is odd or even, then \((s_{1} + s_{2} + \dots + s_{n})/n \to \frac{1}{2}\) as \(n \to \infty\).

[This example proves that the converse of Ex. 27 is not true: for \(s_{n}\) oscillates as \(n \to \infty\).]


30. If \(c_{n}\), \(s_{n}\) denote the sums of the first \(n\) terms of the series \[\tfrac{1}{2} + \cos\theta + \cos 2\theta + \dots,\quad \sin\theta + \sin 2\theta + \dots,\] then \[\lim (c_{1} + c_{2} + \dots + c_{n})/n = 0,\quad \lim (s_{1} + s_{2} + \dots + s_{n})/n = \tfrac{1}{2} \cot\tfrac{1}{2} \theta.\]

  1. A proof that \(\lim\{\psi(n) – \phi(n)\} = 0\), and that therefore each function tends to the limit \(e\), will be found in Chrystal’s Algebra, vol. ii, p. 78. We shall however prove this in Ch. IX by a different method.↩︎