1. Discuss the convergence of the series \(\sum n^{k}\{\sqrt{n + 1} – 2\sqrt{n} + \sqrt{n – 1}\}\), where \(k\) is real.

2. Show that \[\sum n^{r} \Delta^{k}(n^{s}),\] where \[\Delta u_{n} = u_{n} – u_{n+1},\quad \Delta^{2} u_{n} = \Delta(\Delta u_{n}),\] and so on, is convergent if and only if \(k > r + s + 1\), except when \(s\) is a positive integer less than \(k\), when every term of the series is zero.

[The result of Ch.VII, Misc. Ex. 11, shows that \(\Delta^{k}(n^{s})\) is in general of order \(n^{s-k}\).]

3. Show that \[\sum_{1}^{\infty} \frac{n^{2} + 9n + 5}{(n + 1)(2n + 3)(2n + 5)(n + 4)} = \frac{5}{36}.\]

[Resolve the general term into partial fractions.]

4. Show that, if \(R(n)\) is any rational function of \(n\), we can determine a polynomial \(P(n)\) and a constant \(A\) such that \(\sum \{R(n) – P(n) – (A/n)\}\) is convergent. Consider in particular the cases in which \(R(n)\) is one of the functions \(1/(an + b)\), \((an^{2} + 2bn + c)/(\alpha n^{2} + 2\beta n + \gamma)\).

5. Show that the series \[1 – \frac{1}{1 + z} + \frac{1}{2} – \frac{1}{2 + z} + \frac{1}{3} – \frac{1}{3 + z} + \dots\] is convergent provided only that \(z\) is not a negative integer.

6. Investigate the convergence or divergence of the series \[\begin{gathered} \sum \sin\frac{a}{n},\quad \sum \frac{1}{n} \sin\frac{a}{n},\quad \sum (-1)^{n} \sin\frac{a}{n},\\ \sum \left(1 – \cos\frac{a}{n}\right),\quad \sum (-1)^{n} n\left(1 – \cos\frac{a}{n}\right),\end{gathered}\] where \(a\) is real.

7. Discuss the convergence of the series \[\sum_{1}^{\infty} \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right) \frac{\sin(n\theta + \alpha)}{n},\] where \(\theta\) and \(\alpha\) are real.

8. Prove that the series \[1 – \tfrac{1}{2} – \tfrac{1}{3} + \tfrac{1}{4} + \tfrac{1}{5} + \tfrac{1}{6} – \tfrac{1}{7} – \tfrac{1}{8} – \tfrac{1}{9} – \tfrac{1}{10} + \dots,\] in which successive terms of the same sign form groups of \(1\), \(2\), \(3\), \(4\), … terms, is convergent; but that the corresponding series in which the groups contain \(1\), \(2\), \(4\), \(8\), … terms oscillates finitely.

9. If \(u_{1}\), \(u_{2}\), \(u_{3}\), … is a decreasing sequence of positive numbers whose limit is zero, then the series \[u_{1} – \tfrac{1}{2}(u_{1} + u_{2}) + \tfrac{1}{3}(u_{1} + u_{2} + u_{3}) – \dots, \quad u_{1} – \tfrac{1}{3}(u_{1} + u_{3}) + \tfrac{1}{5}(u_{1} + u_{3} + u_{5}) – \dots\] are convergent. [For if \((u_{1} + u_{2} + \dots + u_{n})/n = v_{n}\) then \(v_{1}\), \(v_{2}\), \(v_{3}\), … is also a decreasing sequence whose limit is zero (Ch. IV, Misc. Ex. 8, 27). This shows that the first series is convergent; the second we leave to the reader. In particular the series \[1 – \tfrac{1}{2}\left(1 + \tfrac{1}{2}\right) + \tfrac{1}{3}\left(1 + \tfrac{1}{2} + \tfrac{1}{3}\right) – \dots,\quad 1 – \tfrac{1}{3}\left(1 + \tfrac{1}{{3}}\right) + \tfrac{1}{5}\left(1 + \tfrac{1}{3} + \tfrac{1}{5}\right) – \dots\] are convergent.]

10. If \(u_{0} + u_{1} + u_{2} + \dots\) is a divergent series of positive and decreasing terms, then \[(u_{0} + u_{2} + \dots + u_{2n})/(u_{1} + u_{3} + \dots + u_{2n+1}) \to 1.\]

11. Prove that if \(\alpha > 0\) then \(\lim\limits_{p\to\infty} \sum\limits_{n=0}^{\infty} (p + n)^{-1-\alpha} = 0\).

12. Prove that \(\lim\limits_{\alpha\to 0+} \alpha \sum\limits_{1}^{\infty} n^{-1-\alpha} = 1\). [It follows from § 174 that \[0 < 1^{-1-\alpha} + 2^{-1-\alpha} + \dots + (n – 1)^{-1-\alpha} – \int_{1}^{n} x^{-1-\alpha}\, dx \leq 1,\] and it is easy to deduce that \(\sum n^{-1-\alpha}\) lies between \(1/\alpha\) and \((1/\alpha) + 1\).]

13. Find the sum of the series \(\sum\limits_{1}^{\infty} u_{n}\), where \[u_{n} = \frac{x^{n} – x^{-n-1}}{(x^{n} + x^{-n})(x^{n+1} + x^{-n-1}) } = \frac{1}{x – 1} \left(\frac{1}{x^{n} + x^{-n}} – \frac{1}{x^{n+1} + x^{-n-1}}\right),\] for all real values of \(x\) for which the series is convergent.

[If \(|x|\) is not equal to unity then the series has the sum \(x/\{(x – 1)(x^{2} + 1)\}\). If \(x = 1\) then \(u_{n} = 0\) and the sum is \(0\). If \(x = -1\) then \(u_{n} = \frac{1}{2}(-1)^{n+1}\) and the series oscillates finitely.]

14. Find the sums of the series \[\frac{z}{1 + z} + \frac{2z^{2}}{1 + z^{2}} + \frac{4z^{4}}{1 + z^{4}} + \dots,\quad \frac{z}{1 – z^{2}} + \frac{z^{2}}{1 – z^{4}} + \frac{z^{4}}{1 – z^{8}} + \dots\] (in which all the indices are powers of \(2\)), whenever they are convergent.

[The first series converges only if \(|z| < 1\), its sum then being \(z/(1 – z)\); the second series converges to \(z/(1 – z)\) if \(|z| < 1\) and to \(1/(1 – z)\) if \(|z| > 1\).]

15. If \(|a_{n}| \leq 1\) for all values of \(n\) then the equation \[0 = 1 + a_{1}z + a_{2}z^{2} + \dots\] cannot have a root whose modulus is less than \(\frac{1}{2}\), and the only case in which it can have a root whose modulus is equal to \(\frac{1}{2}\) is that in which \(a_{n} = -\operatorname{Cis}(n\theta)\), when \(z = \frac{1}{2} \operatorname{Cis}(-\theta)\) is a root.

16. **Recurring Series.** A power series \(\sum a_{n}z^{n}\) is said to be a *recurring series* if its coefficients satisfy a relation of the type \[\begin{equation*} a_{n} + p_{1}a_{n-1} + p_{2}a_{n-2} + \dots + p_{k}a_{n-k} = 0, \tag{1} \end{equation*}\] where \(n \geq k\) and \(p_{1}\), \(p_{2}\), …, \(p_{k}\) are independent of \(n\). Any recurring series is the expansion of a rational function of \(z\). To prove this we observe in the first place that the series is certainly convergent for values of \(z\) whose modulus is sufficiently small. For let \(G\) be the greater of the two numbers \[1,\quad |p_{1}| + |p_{2}| + \dots + |p_{k}|.\] Then it follows from the equation (1) that \(|a_{n}| \leq G\alpha_{n}\), where \(\alpha_{n}\) is the modulus of the numerically greatest of the preceding coefficients; and from this that \(|a_{n}| < KG^{n}\), where \(K\) is independent of \(n\). Thus the recurring series is certainly convergent for values of \(z\) whose modulus is less than \(1/G\).

But if we multiply the series \(f(z) = \sum a_{n}z^{n}\) by \(p_{1}z\), \(p_{2}z^{2}\), …, \(p_{k}z^{k}\), and add the results, we obtain a new series in which all the coefficients after the \((k – 1)\)th vanish in virtue of the relation (1), so that \[(1 + p_{1}z + p_{2}z^{2} + \dots + p_{k}z^{k})f(z) = P_{0} + P_{1}z + \dots + P_{k-1}z^{k-1},\] where \(P_{0}\), \(P_{1}\), …, \(P_{k-1}\) are constants. The polynomial \(1 + p_{1}z + p_{2}z^{2} + \dots + p_{k}z^{k}\) is called the *scale of relation* of the series.

Conversely, it follows from the known results as to the expression of any rational function as the sum of a polynomial and certain partial fractions of the type \(A/(z – a)^{p}\), and from the Binomial Theorem for a negative integral exponent, that any rational function whose denominator is not divisible by \(z\) can be expanded in a power series convergent for values of \(z\) whose modulus is sufficiently small, in fact if \(|z| < \rho\), where \(\rho\) is the least of the moduli of the roots of the denominator (cf. Ch. IV, Misc. Ex. 18 *et seq.*). And it is easy to see, by reversing the argument above, that the series is a recurring series. Thus

the necessary and sufficient condition that a power series should be a recurring series is that it should be the expansion of such a rational function of \(z\).

17. **Solution of Difference-Equations.** A relation of the type of (1) in Ex. 16 is called a *linear difference-equation in \(a_{n}\) with constant coefficients*. Such equations may be solved by a method which will be sufficiently explained by an example. Suppose that the equation is \[a_{n} – a_{n-1} – 8a_{n-2} + 12a_{n-3} = 0.\] Consider the recurring power series \(\sum a_{n}z^{n}\). We find, as in Ex. 16, that its sum is \[\frac{a_{0} + (a_{1} – a_{0}) z + (a_{2} – a_{1} – 8a_{0}) z^{2}} {1 – z – 8z^{2} + 12z^{3}} = \frac{A_{1}}{1 – 2z} + \frac{A_{2}}{(1 – 2z)^{2}} + \frac{B}{1 + 3z},\] where \(A_{1}\), \(A_{2}\), and \(B\) are numbers easily expressible in terms of \(a_{0}\), \(a_{1}\), and \(a_{2}\). Expanding each fraction separately we see that the coefficient of \(z^{n}\) is \[a_{n} = 2^{n}\{A_{1} + (n + 1) A_{2}\} + (-3)^{n} B.\] The values of \(A_{1}\), \(A_{2}\), \(B\) depend upon the first three coefficients \(a_{0}\), \(a_{1}\), \(a_{2}\), which may of course be chosen arbitrarily.

18. The solution of the difference-equation \(u_{n} – 2\cos\theta u_{n-1} + u_{n-2} = 0\) is \(u_{n} = A\cos n\theta + B\sin n\theta\), where \(A\) and \(B\) are arbitrary constants.

19. If \(u_{n}\) is a polynomial in \(n\) of degree \(k\), then \(\sum u_{n} z^{n}\) is a recurring series whose scale of relation is \((1 – z)^{k+1}\).

20. Expand \(9/\{(z – 1)(z + 2)^{2}\}\) in ascending powers of \(z\).

21. Prove that if \(f(n)\) is the coefficient of \(z^{n}\) in the expansion of \(z/(1 + z + z^{2})\) in powers of \(z\), then \[(1)\ f(n) + f(n – 1) + f(n – 2) = 0,\quad (2)\ f(n) = (\omega_{3}^{n} – \omega_{3}^{2n})/(\omega_{3} – \omega_{3}^{2}),\] where \(\omega_{3}\) is a complex cube root of unity. Deduce that \(f(n)\) is equal to \(0\) or \(1\) or \(-1\) according as \(n\) is of the form \(3k\) or \(3k + 1\) or \(3k + 2\), and verify this by means of the identity \(z/(1 + z + z^{2}) = z(1 – z)/(1 – z^{3})\).

22. A player tossing a coin is to score one point for every head he turns up and two for every tail, and is to play on until his score reaches or passes a total \(n\). Show that his chance of making exactly the total \(n\) is \(\frac{1}{3}\{2 + (-\frac{1}{2})^{n}\}\).

[If \(p_{n}\) is the probability then \(p_{n} = \frac{1}{2} (p_{n-1} + p_{n-2})\). Also \(p_{0} = 1\), \(p_{1} = \frac{1}{2}\).]

23. Prove that \[\frac{1}{a + 1} + \frac{1}{a + 2} + \dots + \frac{1}{a + n} = \binom{n}{1}\frac{1}{a + 1} – \binom{n}{2}\frac{1!}{(a + 1)(a + 2)} + \dots\] if \(n\) is a positive integer and \(a\) is not one of the numbers \(-1\), \(-2\), …, \(-n\).

[This follows from splitting up each term on the right-hand side into partial fractions. When \(a > -1\), the result may be deduced very simply from the equation \[\int_{0}^{1} x^{a}\frac{1 – x^{n}}{1 – x}\, dx = \int_{0}^{1} (1 – x)^{a}\{1 – (1 – x)^{n}\}\frac{dx}{x}\] by expanding \((1 – x^{n})/(1 – x)\) and \(1 – (1 – x)^{n}\) in powers of \(x\) and integrating each term separately. The result, being merely an algebraical identity, must be true for all values of \(a\) save \(-1\), \(-2\), …, \(-n\).]

24. Prove by multiplication of series that \[\sum_{0}^{\infty} \frac{z^{n}}{n!} \sum_{1}^{\infty} \frac{(-1)^{n-1}z^{n}}{n\cdot n!} = \sum_{1}^{\infty} \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right) \frac{z^{n}}{n!}.\]

[The coefficient of \(z^{n}\) will be found to be \[\frac{1}{n!}\left\{ \binom{n}{1} – \frac{1}{2}\binom{n}{2} + \frac{1}{3}\binom{n}{3} – \dots \right\}.\] Now use Ex. 23, taking \(a = 0\).]

25. If \(A_{n} \to A\) and \(B_{n} \to B\) as \(n \to \infty\), then \[(A_{1}B_{n} + A_{2}B_{n-1} + \dots + A_{n}B_{1})/n \to AB.\]

[Let \(A_{n} = A + \epsilon_{n}\). Then the expression given is equal to \[A \frac{B_{1} + B_{2} + \dots + B_{n}}{n} + \frac{\epsilon_{1}B_{n} + \epsilon_{2}B_{n-1} + \dots + \epsilon_{n}B_{1}}{n}.\]The first term tends to \(AB\) (Ch. IV, Misc. Ex. 27). The modulus of the second is less than \(\beta\{|\epsilon_{1}| + |\epsilon_{2}| + \dots + |\epsilon_{n}|\}/n\), where \(\beta\) is any number greater than the greatest value of \(|B_{\nu}|\): and this expression tends to zero.]

26. Prove that if \(c_{n} = a_{1}b_{n} + a_{2}b_{n-1} + \dots + a_{n}b_{1}\) and \[A_{n} = a_{1} + a_{2} + \dots + a_{n},\quad B_{n} = b_{1} + b_{2} + \dots + b_{n},\quad C_{n} = c_{1} + c_{2} + \dots + c_{n},\] then \[C_{n} = a_{1}B_{n} + a_{2}B_{n-1} + \dots + a_{n}B_{1} = b_{1}A_{n} + b_{2}A_{n-1} + \dots + b_{n}A_{1}\] and \[C_{1} + C_{2} + \dots + C_{n} = A_{1}B_{n} + A_{2}B_{n-1} + \dots + A_{n}B_{1}.\]

Hence prove that if the series \(\sum a_{n}\), \(\sum b_{n}\) are convergent and have the sums \(A\), \(B\), so that \(A_{n} \to A\), \(B_{n} \to B\), then \[(C_{1} + C_{2} + \dots + C_{n})/n \to AB.\] Deduce that *if \(\sum c_{n}\) is convergent then its sum is \(AB\)*. This result is known as ** Abel’s Theorem on the multiplication of Series**. We have already seen that we can multiply the series \(\sum a_{n}\), \(\sum b_{n}\) in this way if both series are

*absolutely*convergent: Abel’s Theorem shows that we can do so even if one or both are not absolutely convergent,

*provided only that the product series is convergent*.

27. Prove that \[\begin{aligned} \tfrac{1}{2} \left(1 – \tfrac{1}{2} + \tfrac{1}{3} – \dots\right)^{2} &= \tfrac{1}{2} – \tfrac{1}{3} \left(1 + \tfrac{1}{2}\right) + \tfrac{1}{4} \left(1 + \tfrac{1}{2} + \tfrac{1}{3}\right) – \dots,\\ \tfrac{1}{2} \left(1 – \tfrac{1}{3} + \tfrac{1}{5} – \dots\right)^{2} &= \tfrac{1}{2} – \tfrac{1}{4} \left(1 + \tfrac{1}{3}\right) + \tfrac{1}{6} \left(1 + \tfrac{1}{3} + \tfrac{1}{5}\right) – \dots.\end{aligned}\]

[Use Ex. 9 to establish the convergence of the series.]

28. For what values of \(m\) and \(n\) is the integral \(\int_{0}^{\pi} \sin^{m} x (1 – \cos x)^{n}\, dx\) convergent? [If \(m + 1\) and \(m + 2n + 1\) are positive.]

29. Prove that if \(a > 1\) then \[\int_{-1}^{1} \frac{dx}{(a – x) \sqrt{1 – x^{2}}} = \frac{\pi}{\sqrt{a^{2} – 1}}.\]

30. Establish the formulae \[\begin{aligned} {2} \int_{0}^{\infty} F\{\sqrt{x^{2} + 1} + x\}\, dx &= \tfrac{1}{2}\int_{1}^{\infty} &&\left(1 + \frac{1}{y^{2}}\right) F(y) \, dy,\\ \int_{0}^{\infty} F\{\sqrt{x^{2} + 1} – x\}\, dx &= \tfrac{1}{2}\int_{0}^{1} &&\left(1 + \frac{1}{y^{2}}\right) F(y)\, dy.\end{aligned}\] In particular, prove that if \(n > 1\) then \[\int_{0}^{\infty} \frac{dx}{\{\sqrt{x^{2} + 1} + x\}^{n}} = \int_{0}^{\infty} \{\sqrt{x^{2} + 1} – x\}^{n}\, dx = \frac{n}{n^{2} – 1}.\]

[In this and the succeeding examples it is of course supposed that the arbitrary functions which occur are such that the integrals considered have a meaning in accordance with the definitions of § 177*et seq.*]

31. Show that if \(2y = ax – (b/x)\), where \(a\) and \(b\) are positive, then \(y\) increases steadily from \(-\infty\) to \(\infty\) as \(x\) increases from \(0\) to \(\infty\). Hence show that \[\begin{aligned} \int_{0}^{\infty} f\left\{\tfrac{1}{2}\left(ax + \frac{b}{x}\right)\right\} dx &= \frac{1}{a} \int_{-\infty}^{\infty} f\{\sqrt{y^{2} + ab}\} \left\{1 + \frac{y}{\sqrt{y^{2} + ab}}\right\} dy\\ &= \frac{2}{a} \int_{0}^{\infty} f\{\sqrt{y^{2} + ab}\}\, dy.\end{aligned}\]

32. Show that if \(2y = ax + (b/x)\), where \(a\) and \(b\) are positive, then two values of \(x\) correspond to any value of \(y\) greater than \(\sqrt{ab}\). Denoting the greater of these by \(x_{1}\) and the less by \(x_{2}\), show that, as \(y\) increases from \(\sqrt{ab}\) towards \(\infty\), \(x_{1}\) increases from \(\sqrt{b/a}\) towards \(\infty\), and \(x_{2}\) decreases from \(\sqrt{b/a}\) to \(0\). Hence show that \[\begin{aligned} \int_{\sqrt{b/a}}^{\infty} f(y)\, dx_{1} &= \frac{1}{a} \int_{\sqrt{ab}}^{\infty} f(y) \left\{\frac{y}{\sqrt{y^{2} – ab}} + 1\right\} dy,\\ \int_{0}^{\sqrt{b/a}} f(y)\, dx_{2} &= \frac{1}{a} \int_{\sqrt{ab}}^{\infty} f(y) \left\{\frac{y}{\sqrt{y^{2} – ab}} – 1\right\} dy,\end{aligned}\] and that \[\int_{0}^{\infty} f\left\{\tfrac{1}{2}\left(ax + \frac{b}{x}\right)\right\} dx = \frac{2}{a} \int_{\sqrt{ab}}^{\infty} \frac{yf(y)}{\sqrt{y^{2} – ab}}\, dy = \frac{2}{a} \int_{0}^{\infty} f\{\sqrt{z^{2} + ab}\}\, dz.\]

33. Prove the formula \[\int_{0}^{\pi} f(\sec\tfrac{1}{2}x + \tan\tfrac{1}{2}x)\frac{dx}{\sqrt{\sin x}} = \int_{0}^{\pi} f(\operatorname{cosec} x)\frac{dx}{\sqrt{\sin x}}.\]

34. If \(a\) and \(b\) are positive, then \[\int_{0}^{\infty} \frac{dx}{(x^{2} + a^{2})(x^{2} + b^{2})} = \frac{\pi}{2ab(a + b)},\quad \int_{0}^{\infty} \frac{x^{2}\, dx}{(x^{2} + a^{2})(x^{2} + b^{2})} = \frac{\pi}{2(a + b)}.\] Deduce that if \(\alpha\), \(\beta\), and \(\gamma\) are positive, and \(\beta^{2} \geq \alpha\gamma\), then \[\int_{0}^{\infty} \frac{dx}{\alpha x^{4} + 2\beta x^{2} + \gamma} = \frac{\pi}{2\sqrt{2\gamma A}}, \quad \int_{0}^{\infty} \frac{x^{2}\, dx}{\alpha x^{4} + 2\beta x^{2} + \gamma} = \frac{\pi}{2\sqrt{2\alpha A}},\] where \(A = \beta + \sqrt{\alpha\gamma}\). Also deduce the last result from Ex. 31, by putting \(f(y) = 1/(c^{2} + y^{2})\). The last two results remain true when \(\beta^{2} < \alpha\gamma\), but their proof is then not quite so simple.

35. Prove that if \(b\) is positive then \[\int_{0}^{\infty} \frac{x^{2}\, dx}{(x^{2} – a^{2})^{2} + b^{2}x^{2}} = \frac{\pi}{2b},\quad \int_{0}^{\infty} \frac{x^{4}\, dx}{\{(x^{2} – a^{2})^{2} + b^{2}x^{2}\}^{2}} = \frac{\pi}{4b^{3}}.\]

36. Extend Schwarz’s inequality (Ch. VII, Misc. Ex. 42) to infinite integrals of the first and second kinds.

37. Prove that if \(\phi(x)\) is the function considered at the end of § 178 then \[\int_{0}^{\infty} \phi(x)\, dx = \sum_{0}^{\infty} \frac{1}{(n + 1)^{2}}.\]

38. Prove that \[\begin{aligned} {2} \int_{1}^{\infty} dx \left(\int_{1}^{\infty} \frac{x – y}{(x + y)^{3}}\, dy\right) &= -1, \quad & \int_{1}^{\infty} dy \left(\int_{1}^{\infty} \frac{x – y}{(x + y)^{3}}\, dx\right) &= 1;\\ \int_{1}^{\infty} dx \left(\int_{1}^{\infty} \frac{x^{2} – y^{2}}{(x^{2} + y^{2})^{2}}\, dy\right) &= -\tfrac{1}{4}\pi, \quad & \int_{1}^{\infty} dy \left(\int_{1}^{\infty} \frac{x^{2} – y^{2}}{(x^{2} + y^{2})^{2}}\, dx\right) &= \tfrac{1}{4}\pi.\end{aligned}\]

Establish similar results in which the limits of integration are \(0\) and \(1\).

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