Sequences and series, recurrence and convergence

1. The sequence \(x_n\) for \(n = 0, 1, 2, \ldots\) is defined by \(x_0 = 1\) and by \[x_{n+1} = \frac{x_n + 2}{x_n + 1}\] for \(n \geqslant 0\).
   1. Explain briefly why \(x_n \geqslant 1\) for all \(n\).
   2. Show that \(x_{n+1}^2 - 2\) and \(x_n^2 - 2\) have opposite sign, and that \[\left|x_{n+1}^2 - 2\right| \leqslant \tfrac{1}{4}\left|x_n^2 - 2\right|\,.\]
   3. Show that \[2 - 10^{-6} \leqslant x_{10}^2 \leqslant 2\,.\]
2. The sequence \(y_n\) for \(n = 0, 1, 2, \ldots\) is defined by \(y_0 = 1\) and by \[y_{n+1} = \frac{y_n^2 + 2}{2y_n}\] for \(n \geqslant 0\).
   1. Show that, for \(n \geqslant 0\), \[y_{n+1} - \sqrt{2} = \frac{(y_n - \sqrt{2})^2}{2y_n}\] and deduce that \(y_n \geqslant 1\) for \(n \geqslant 0\).
   2. Show that \[y_n - \sqrt{2} \leqslant 2\left(\frac{\sqrt{2}-1}{2}\right)^{2^n}\] for \(n \geqslant 1\).
   3. Using the fact that \[\sqrt{2} - 1 < \tfrac{1}{2}\,,\] or otherwise, show that \[\sqrt{2} \leqslant y_{10} \leqslant \sqrt{2} + 10^{-600}\,.\]

A sequence \(x_1, x_2, \ldots\) of real numbers is defined by \(x_{n+1} = x_n^2 - 2\) for \(n \geqslant 1\) and \(x_1 = a\).

1. Show that if \(a > 2\) then \(x_n \geqslant 2 + 4^{n-1}(a-2)\).
2. Show also that \(x_n \to \infty\) as \(n \to \infty\) if and only if \(|a| > 2\).
3. When \(a > 2\), a second sequence \(y_1, y_2, \ldots\) is defined by \[ y_n = \frac{Ax_1 x_2 \cdots x_n}{x_{n+1}}, \] where \(A\) is a positive constant and \(n \geqslant 1\). Prove that, for a certain value of \(a\), with \(a > 2\), which you should find in terms of \(A\), \[ y_n = \frac{\sqrt{x_{n+1}^2 - 4}}{x_{n+1}} \] for all \(n \geqslant 1\). Determine whether, for this value of \(a\), the second sequence converges.

1. Prove that, for any positive integers \(n\) and \(r\), \[ \frac{1}{^{n+r}\C_{r+1}} =\frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right). \] Hence determine \[ \sum_{n=1}^{\infty}{\frac{1}{^{n+r}\C_{r+1}}} \,, \] and deduce that \ \(\displaystyle \sum_{n=2}^\infty \frac 1 {^{n+2}\C_3} = \frac12\,\).
2. Show that, for \(n \ge 3\,\), \[ \frac{3!}{n^3} < \frac{1}{^{n+1}\C_{3}} \ \ \ \ \ \text{and} \ \ \ \ \ \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} < \frac{5!}{n^3} \,. \] By summing these inequalities for \(n \ge 3\,\), show that \[ \frac{115}{96} < \sum_{n=1}^{\infty}{\frac{1}{n^3}} < \frac{116}{96} \, . \]

Prove that, for any numbers \(a_1, a_2, \ldots\,,\) and \(b_1, b_2, \ldots\,,\) and for \(n\ge1\), \[ \sum_{m=1}^n a_m(b_{m+1} -b_m) = a_{n+1}b_{n+1} -a_1b_1 -\sum_{m=1}^n b_{m+1}(a_{m+1} -a_m) \,. \]

1. By setting \(b_m = \sin mx\), show that \[ \sum_{m=1}^n \cos (m+\tfrac12)x = \tfrac12 \big(\sin (n+1)x - \sin x \big) \cosec \tfrac12 x \,. \] Note: $\sin A - \sin B = \displaystyle 2 \cos \big( \tfrac{{\displaystyle A+B\vphantom{_1}}} {\displaystyle 2\vphantom{^1}} \big)\, \sin\big( \tfrac{{\displaystyle A-B\vphantom{_1}}}{\displaystyle 2\vphantom{^1}} \big)\, $.
2. Show that \[ \sum_{m=1}^n m\sin mx = \big (p \sin(n+1)x +q \sin nx\big) \cosec^2 \tfrac12 x \,, \] where \(p\) and \(q\) are to be determined in terms of \(n\). Note: \(2\sin A \sin B = \cos (A-B) - \cos (A+B)\,\); Note: \(2\cos A \sin B = \sin (A+B) - \sin (A-B)\,\).

Evaluate the integral \[ \hphantom{ \ \ \ \ \ \ \ \ \ (m> \tfrac12)\,.} \int_{m-\frac12} ^\infty \frac 1{x^2}\, \d x { \ \ \ \ \ \ \ \ \ (m > \tfrac12)\,.} \] Show by means of a sketch that \[ \sum_{r=m}^n \frac 1 {r^2} \approx \int_{m-\frac12}^{n+\frac12} \frac1 {x^2} \, \d x \,, \tag{\(*\)} \] where \(m\) and \(n\) are positive integers with \(m < n\).

1. You are given that the infinite series \(\displaystyle \sum_{r=1}^\infty \frac 1 {r^2}\) converges to a value denoted by \(E\). Use \((*)\) to obtain the following approximations for \(E\): \[ E\approx 2\,; \ \ \ \ E\approx \frac53\,; \ \ \ \ E\approx \frac{33}{20} \,.\]
2. Show that, when \(r\) is large, the error in approximating \(\dfrac 1{r^2}\) by \(\displaystyle \int_{r-\frac12}^{r+\frac12} \frac 1 {x^2} \, \d x\) is approximately \(\dfrac 1{4r^4}\,\). Given that \(E \approx 1.645\), show that \(\displaystyle \sum_{r=1}^\infty \frac1{r^4} \approx 1.08\, \).

Show Solution

\begin{align*} && \int_{m-\frac12}^\infty \frac{1}{x^2} \d x &= \lim_{K \to \infty} \left [ -x^{-1} \right]_{m-\frac12}^K \\ &&&= \frac{1}{m-\frac12} - \lim_{K \to \infty }\frac{1}K \\ &&&= \frac{1}{m-\frac12} \end{align*}

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Notice that \(\displaystyle \frac{1}{r^2} \approx \int_{r-\frac12}^{r+\frac12} \frac{1}{x^2} \d x\) as the area of the orange boxes and under the blue lines are similar.

1. \(\,\) \begin{align*} E &\approx \int_{1-\frac12}^\infty \frac1{x^2} \d x = \frac{1}{1-\frac12} = 2 \\ E &\approx 1 + \int_{2-\frac12}^\infty \frac1{x^2} \d x= 1 + \frac{1}{2 - \frac12} = \frac53 \\ E &\approx 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x= \frac54 + \frac{1}{3-\frac12} \\ &= \frac54+\frac{2}{5} = \frac{33}{20} \end{align*}
2. The error is \begin{align*} && \epsilon &= \int_{r-\frac12}^{r+\frac12} \frac 1 {x^2} \, \d x - \frac1{r^2} \\ &&&= \frac{1}{r-\frac12} - \frac{1}{r + \frac12} - \frac1{r^2} \\ &&&= \frac{1}{r^2 - \frac14} - \frac1{r^2} \\ &&&= \frac{\frac14}{r^2(r^2-\frac14)} \\ &&&\approx \frac{1}{4r^4} \end{align*} Therefore \begin{align*} && \sum_{n=1}^\infty \frac1{r^4} &\approx 4 \left ( 1 +\frac14 + \int_{3-\frac12}^\infty \frac1{x^2} \d x-\sum_{r=1}^\infty \frac{1}{r^2} \right) + 1 + \frac{1}{2^4}\\ &&&= 4 \left ( \frac{33}{20}-1.645 \right) + 1 + \frac{1}{2^4} \\ &&&= 4 \left ( 1.65-1.645 \right) + 1 + \frac{1}{2^4} \\ &&&= 1.0825 \approx 1.08 \end{align*}

1. By considering \(\displaystyle \frac1{1+ x^r} - \frac1{1+ x^{r +1}}\) for \(\vert x \vert \ne 1\), simplify \[ \sum_{r=1}^N \frac{x^r}{(1+x^r)(1+x^{r+1})} \] Show that, for \(\vert x \vert <1\), \[ \sum_{r=1}^\infty \frac{x^r}{(1+x^r)(1+x^{r+1})} = \frac x {1-x^2} \]
2. Deduce that \[ \sum_{r=1}^\infty \textrm{sech}(ry)\textrm{sech}((r + 1)y) = 2\e^{-y} \textrm{cosech}(2 y) \] for \(y > 0\). Hence simplify \[ \sum_{r=-\infty}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) \] for \(y>0\).

The numbers \(f(r)\) satisfy \(f(r)>f(r+1)\) for $r=1, 2, \dots\(. Show that, for any non-negative integer \)n$, \[ k^n(k-1) \, f(k^{n+1}) \le \sum_{r=k^n}^{k^{n+1}-1}f(r) \le k^n(k-1)\, f(k^n)\, \] where \(k\) is an integer greater than 1.

1. By taking \(f(r) = 1/r\), show that \[ \frac{N+1}2 \le \sum_{r=1}^{2^{N+1}-1} \frac1r \le N+1 \,. \] Deduce that the sum \(\displaystyle \sum_{r=1}^\infty \frac1r\) does not converge.
2. By taking \(f(r)= 1/r^3\), show that \[ \sum_{r=1}^\infty \frac1 {r^3} \le 1 \tfrac 13 \,. \]
3. Let \(S(n)\) be the set of positive integers less than \(n\) which do not have a \(2\) in their decimal representation and let \(\sigma(n)\) be the sum of the reciprocals of the numbers in \(S(n)\), so for example \(\sigma(5) = 1+\frac13+\frac14\). Show that \(S(1000)\) contains \(9^3-1\) distinct numbers. Show that \(\sigma (n) < 80\) for all \(n\).

In this question, the following theorem may be used. Let \(u_1\), \(u_2\), \(\ldots\) be a sequence of (real) numbers. If the sequence is bounded above (that is, \(u_n\le b\) for all \(n\), where \(b\) is some fixed number) and increasing (that is, \(u_n \ge u_{n-1}\) for all \(n\)), then the sequence tends to a limit (that is, converges). The sequence \(u_1\), \(u_2\), \(\ldots\) is defined by \(u_1=1\) and \[ u_{n+1} = 1+\frac 1{u_n} \ \ \ \ \ \ \ \ \ \ (n\ge1)\,. \tag{\(*\)} \]

1. Show that, for \(n\ge3\), \[ u_{n+2}-u_n = \frac{u_{n} - u_{n-2}}{(1+u_n)(1+u_{n-2})} . \]
2. Prove, by induction or otherwise, that \(1\le u_n \le 2\) for all \(n\).
3. Show that the sequence \(u_1\), \(u_3\), \(u_5\), \(\ldots\) tends to a limit, and that the sequence \(u_2\), \(u_4\), \(u_6\), \(\ldots\) tends to a limit. Find these limits and deduce that the sequence \(u_1\), \(u_2\), \(u_3\), \(\ldots\,\) tends to a limit. Would this conclusion change if the sequence were defined by \((*)\) and \(u_1=3\)?

A sequence of numbers \(t_0\), \(t_1\), \(t_2\), \(\ldots\,\) satisfies \[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t_{n+2} = p t_{n+1}+qt_{n} \ \ \ \ \ \ \ \ \ \ (n\ge0), \] where \(p\) and \(q\) are real. Throughout this question, \(x\), \(y\) and \(z\) are non-zero real numbers.

1. Show that, if \(t_n=x\) for all values of \(n\), then \(p+q=1\) and \(x\) can be any (non-zero) real number.
2. Show that, if \(t_{2n} = x\) and \(t_{2n+1}=y\) for all values of \(n\), then \(q\pm p=1\). Deduce that either \(x=y\) or \(x=-y\), unless \(p\) and \(q\) take certain values that you should identify.
3. Show that, if \(t_{3n} = x\), \(t_{3n+1}=y\) and \(t_{3n+2}=z\) for all values of \(n\), then \[ p^3+q^3 +3pq-1=0\,. \] Deduce that either \(p+q=1\) or \((p-q)^2 +(p+1)^2+(q+1)^2=0\). Hence show that either \(x=y=z\) or \(x+y+z=0\).

The positive numbers \(\alpha\), \(\beta\) and \(q\) satisfy \(\beta-\alpha >q\). Show that \[ \frac{\alpha^2+\beta^2 -q^2}{\alpha\beta}-2> 0\,. \] The sequence \(u_0\), \(u_1\), \(\ldots\) is defined by \(u_0=\alpha\), \(u_1=\beta\) and \[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u_{n+1} = \frac {u_{n}^2 -q^2}{u_{n-1}} \ \ \ \ \ \ \ \ \ \ \ (n\ge1), \] where \(\alpha\), \(\beta\) and \(q\) are given positive numbers (and \(\alpha\) and \(\beta\) are such that no term in the sequence is zero). Prove that \(u_n(u_n+u_{n+2}) = u_{n+1}(u_{n-1}+u_{n+1})\,\). Prove also that \[ u_{n+1} -pu_n + u_{n-1}=0 \] for some number \(p\) which you should express in terms of \(\alpha\), \(\beta\) and \(q\). Hence, or otherwise, show that if \(\beta> \alpha+q\), the sequence is strictly increasing (that is, \(u_{n+1}-u_n > 0\) for all \(n\)). Comment on the case \(\beta =\alpha +q\).

In this question, \(\vert x \vert <1\) and you may ignore issues of convergence.

1. Simplify \[ (1-x)(1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n})\,, \] where \(n\) is a positive integer, and deduce that \[ \frac1{1-x} = (1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) + \frac {x^{2^{n+1}}}{1-x}\,. \] Deduce further that \[ \ln(1-x) = - \sum_{r=0}^\infty \ln \left (1+ x ^{2^r}\right) \,, \] and hence that \[ \frac1 {1-x} = \frac 1 {1+x} + \frac {2x}{1+x^2} + \frac {4x^3}{1+x^4} +\cdots\,. \]
2. Show that \[ \frac{1+2x}{1+x+x^2} = \frac{1-2x}{1-x+x^2} + \frac{2x-4x^3}{1-x^2+x^4} + \frac {4x^3-8x^7}{1-x^4+x^8} + \cdots\,. \]

The sequence \(F_0\), \(F_1\), \(F_2\), \(\ldots\,\) is defined by \(F_0=0\), \(F_1=1\) and, for \(n\ge0\), \[ F_{n+2} = F_{n+1} + F_n \,. \]

1. Show that \(F_0F_3-F_1F_2 = F_2F_5- F_3F_4\,\).
2. Find the values of \(F_nF_{n+3} - F_{n+1}F_{n+2}\) in the two cases that arise.
3. Prove that, for \(r=1\), \(2\), \(3\), \(\ldots\,\), \[ \arctan \left( \frac 1{F_{2r}}\right) =\arctan \left( \frac 1{F_{2r+1}}\right)+ \arctan \left( \frac 1{F_{2r+2}}\right) \] and hence evaluate the following sum (which you may assume converges): \[ \sum_{r=1}^\infty \arctan \left( \frac 1{F_{2r+1}}\right) \,. \]

Given that \(y = \cos(m \arcsin x)\), for \(\vert x \vert <1\), prove that \[ (1-x^2) \frac {\d^2 y}{\d x^2} -x \frac {\d y}{\d x} +m^2y=0\,. \] Obtain a similar equation relating \(\dfrac{\d^3y}{\d x^3}\,\), \(\dfrac{\d^2y}{\d x^2}\, \) and \(\, \dfrac{\d y}{\d x}\,\), and a similar equation relating \(\dfrac{\d^4y}{\d x^4}\,\), \(\dfrac{\d^3y}{\d x^3}\,\) and \(\,\dfrac{\d^2 y}{\d x^2}\,\). Conjecture and prove a relation between \(\dfrac{\d^{n+2}y}{\d x^{n+2}}\,\), \(\dfrac{\d^{n+1}y}{\d x^{n+1}}\;\) and \(\;\dfrac{\d^n y}{\d x^n}\,\). Obtain the first three non-zero terms of the Maclaurin series for \(y\). Show that, if \(m\) is an even integer, \(\cos m\theta\) may be written as a polynomial in \(\sin\theta\) beginning \[ 1 - \frac{m^2\sin^2\theta}{2!}+ \frac{m^2(m^2-2^2)\sin^4\theta}{4!} -\cdots \,. \, \tag{\(\vert\theta\vert < \tfrac12 \pi\)} \] State the degree of the polynomial.

Let \(S_k(n) \equiv \sum\limits_{r=0}^n r^k\,\), where \(k\) is a positive integer, so that \[ S_1(n) \equiv \tfrac12 n(n+1) \text{ and } S_2(n) \equiv \tfrac16 n(n+1)(2n+1)\,. \]

1. By considering \(\sum\limits_{r=0}^n \left[ (r+1)^k-r^k\right]\, \), show that \[ kS_{k-1}(n)=(n+1)^k -(n+1) - \binom{k}{2} S_{k-2}(n) - \binom {k}{3} S_{k-3}(n) - \cdots - \binom{k}{k-1} S_{1}(n) \;. \tag{\(*\)} \] Obtain simplified expressions for \(S_3(n)\) and \(S_4(n)\).
2. Explain, using \((*)\), why \(S_k(n)\) is a polynomial of degree \(k+1\) in \(n\). Show that in this polynomial the constant term is zero and the sum of the coefficients is 1.

The sequence of real numbers \(u_1\), \(u_2\), \(u_3\), \(\ldots\) is defined by \begin{equation*} u_1=2 \,, \qquad\text{and} \qquad u_{n+1} = k - \frac{36}{u_n} \quad \text{for } n\ge1, \tag{\(*\)} \end{equation*} where \(k\) is a constant.

1. Determine the values of \(k\) for which the sequence \((*)\) is: (a) constant; (b) periodic with period 2; (c) periodic with period 4.
2. In the case \(k=37\), show that \(u_n\ge 2\) for all \(n\). Given that in this case the sequence \((*)\) converges to a limit \(\ell\), find the value of \(\ell\).

The sequence \(u_n\) (\(n= 1, 2, \ldots\)) satisfies the recurrence relation \[ u_{n+2}= \frac{u_{n+1}}{u_n}(ku_n-u_{n+1}) \] where \(k\) is a constant. If \(u_1=a\) and \(u_2=b\,\), where \(a\) and \(b\) are non-zero and \(b \ne ka\,\), prove by induction that \[ u_{2n}=\Big(\frac b a \Big) u_{2n-1} \] \[ u_{2n+1}= c u_{2n} \] for \(n \ge 1\), where \(c\) is a constant to be found in terms of \(k\), \(a\) and \(b\). Hence express \(u_{2n}\) and \(u_{2n-1}\) in terms of \(a\), \(b\), \(c\) and \(n\). Find conditions on \(a\), \(b\) and \(k\) in the three cases:

1. the sequence \(u_n\) is geometric;
2. \(u_n\) has period 2;
3. the sequence \(u_n\) has period 4.

A sequence \(t_0\), \(t_1\), \(t_2\), \(...\) is said to be strictly increasing if \(t_{n+1} > t_n\) for all \(n\ge{0}\,\).

1. The terms of the sequence \(x_0\,\), \(x_1\,\), \(x_2\,\), \(\ldots\) satisfy $$ \ds x_{n+1}=\frac{x_n^2 +6}{5} $$ for \(n\ge{0}\,\). Prove that if \(x_0 > 3\) then the sequence is strictly increasing.
2. The terms of the sequence \(y_0\,\), \(y_1\,\), \(y_2\,\), \(\ldots\) satisfy $$ \ds y_{n+1}= 5-\frac 6 {y_n} $$ for \(n\ge{0}\,\). Prove that if \(2 < y_0 < 3\) then the sequence is strictly increasing but that \(y_n<3\) for all \(n\,\).

Given that \(\f''(x) > 0\) when \(a \le x \le b\,\), explain with the aid of a sketch why \[ (b-a) \, \f \Big( {a+b \over 2} \Big) < \int^b_a \f(x) \, \mathrm{d}x < (b-a) \, \displaystyle \frac{\f(a) + \f(b)}{2} \;. \] By choosing suitable \(a\), \(b\) and \(\f(x)\,\), show that \[ {4 \over (2n-1)^2} < {1 \over n-1} - {1 \over n} < {1 \over 2} \l {1 \over n^2} + {1 \over (n-1)^2}\r \,, \] where \(n\) is an integer greater than 1. Deduce that \[ 4 \l {1 \over 3^2} +{1 \over 5^2} + {1 \over 7^2} + \cdots \r < 1 < {1 \over 2} + \left( {1 \over 2^2} +{1 \over 3^2} + {1 \over 4^2} + \cdots \right)\,. \] Show that \[ {1 \over 2} \l {1 \over 3^2} + {1 \over 4^2} + {1 \over 5^2} + \frac 1 {6^2} + \cdots \r < {1 \over 3^2} +{1 \over 5^2} + {1 \over 7^2} + \cdots \] and hence show that \[ {3 \over 2} \displaystyle < \sum_{n=1}^\infty {1 \over n^2} <{7 \over 4}\;. \]

Given a sequence \(w_0\), \(w_1\), \(w_2\), \(\ldots\,\), the sequence \(F_1\), \(F_2\), \(\ldots\) is defined by $$F_n = w_n^2 + w_{n-1}^2 - 4w_nw_{n-1} \,.$$ Show that $\; F_{n}-F_{n-1} = \l w_n-w_{n-2} \r \l w_n+w_{n-2}-4w_{n-1} \r \; \( for \)n \ge 2\,$.

1. The sequence \(u_0\), \(u_1\), \(u_2\), \(\ldots\) has \(u_0 = 1\), and \(u_1 = 2\) and satisfies \[ u_n = 4u_{n-1} -u_{n-2} \quad (n \ge 2)\;. \] Prove that \ $ u_n^2 + u_{n-1}^2 = 4u_nu_{n-1}-3 \; $ for \(n \ge 1\,\).
2. A sequence \(v_0\), \(v_1\), \(v_2\), \(\ldots\,\) has \(v_0=1\) and satisfies \begin{equation*} v_n^2 + v_{n-1}^2 = 4v_nv_{n-1}-3 \quad (n \ge 1). \tag{\(\ast\)} \end{equation*} \makebox[7mm]{(a) \hfill}Find \(v_1\) and prove that, for each \(n\ge2\,\), either \(v_n= 4v_{n-1} -v_{n-2}\) or \(v_n=v_{n-2}\,\). \makebox[7mm]{(b) \hfill}Show that the sequence, with period 2, defined by \begin{equation*} v_n = \begin{cases} 1 & \mbox{for \(n\) even} \\ 2 & \mbox{for \(n\) odd} \end{cases} \end{equation*} \makebox[7mm]{\hfill}satisfies \((\ast)\). \makebox[7mm]{(c) \hfill}Find a sequence \(v_n\) with period 4 which has \(v_0=1\,\), and satisfies~\((\ast)\).

Show that, if \(n>0\,\), then $$ \int_{e^{1/n}}^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x = {2 \over {n^2\e}}\;. $$ You may assume that \(\ds \frac{\ln x} x \to 0\;\) as \(x\to\infty\,\). Explain why, if \(1 < a < b\,\), then $$ \int_b^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x < \int_a^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x\;. $$ Deduce that $$ \sum_{n=1}^{N}{1 \over n^2} < {\e \over 2}\int_{\e^{1/N}}^{\infty} \left({1-x^{-N}} \over {x^2-x}\right) \ln x\,\d x\;, $$ where \(N\,\) is any integer greater than \(1\).

Show that \[ 2\sin \frac12 \theta \, \cos r\theta = \sin\big(r+\frac12\big)\theta - \sin\big(r-\frac12\big)\theta \;. \] Hence, or otherwise, find all solutions of the equation \[ \cos a\theta + \cos (a + 1) \theta + \dots + \cos(b-2)\theta+\cos (b - 1 ) \theta = 0 \;, \] where \(a\) and \(b\) are positive integers with \(a < b-1\,\).

Given that $$\e = 1 + {1 \over 1 !} + {1 \over 2 !} + {1 \over 3 !} + \cdots + {1 \over r !} + \cdots \; ,$$ use the binomial theorem to show that $$ {\left( 1 + {1 \over n} \right)}^{\!n} < \e $$ for any positive integer \(n\). The product \({\rm P }( n )\) is defined, for any positive integer \(n\), by $$ {\rm P} ( n ) = {3 \over 2} \cdot {5 \over 4} \cdot {9 \over 8} \cdot \ldots \cdot {2^n + 1 \over 2^n} . $$ Use the arithmetic-geometric mean inequality, $$ {a_1 + a_2 + \cdots + a_n \over n} \ge \ {\left( a_1 \cdot a_2 \cdot \ldots \cdot a_n \right)}^{1 \over n}\,, $$ to show that \({\rm P }( n ) < \e\) for all \(n\) . Explain briefly why \({\rm P} ( n )\) tends to a limit as \(n\to\infty\). Show that this limit, \(L\), satisfies \(2 < L\le\e\).

1. Let \(\f(x)=(1+x^2)\e^x\). Show that \(\f'(x)\ge 0\) and sketch the graph of \(\f(x)\). Hence, or otherwise, show that the equation \[ (1+x^2)\e^x = k, \] where \(k\) is a constant, has exactly one real root if \(k>0\) and no real roots if \(k\le 0\).
2. Determine the number of real roots of the equation $$ (\e^x-1) - k \tan^{-1} x=0 $$ in the cases (a) \(0< k\le 2/\pi\) and (b) \(2/\pi < k < 1\).

Justify, by means of a sketch, the formula $$ \lim_{n\rightarrow\infty}\left\{{1\over n}\sum_{m=1}^n \f(1+m/n)\right\} = \int_1^2 \f(x)\,\d x \,. $$ Show that $$ \lim_{n\rightarrow\infty}\left\{{1\over n+1} + {1\over n+2} + \cdots + {1\over n+n}\right\} = \ln 2 \,. $$ Evaluate $$ \lim_{n\rightarrow\infty}\left\{{n\over n^2+1} + {n\over n^2+4} + \cdots + {n\over n^2+n^2}\right\}\,. $$

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\begin{align*} && V &= \lim_{n\rightarrow\infty}\left\{{1\over n+1} + {1\over n+2} + \cdots + {1\over n+n}\right\} \\ && &= \lim_{n\rightarrow\infty}\left\{\sum_{m=1}^n \frac{1}{n+m}\right\} \\ && &= \lim_{n\rightarrow\infty}\left\{\frac1n\sum_{m=1}^n \frac{1}{1+\frac{m}{n}}\right\} \\ &&&=\int_1^2 \frac{1}{x} \d x \\ &&&= \left [\ln x \right]_1^2 = \ln 2 \end{align*} \begin{align*} V &= \lim_{n\rightarrow\infty}\left\{{n\over n^2+1} + {n\over n^2+4} + \cdots + {n\over n^2+n^2}\right\} \\ &= \lim_{n\rightarrow\infty}\left\{\sum_{m=1}^n \frac{n}{n^2+m^2}\right\} \\ &= \lim_{n\rightarrow\infty}\left\{\frac{1}{n}\sum_{m=1}^n \frac{1}{1+\left (\frac{m}{n} \right)^2}\right\} \\ &= \int_0^1 \frac{1}{1+x^2} \d x \\ &= \left [\tan^{-1} x \right]_0^1 \\ &= \frac{\pi}4 \end{align*}

The sequence \(u_0\), \(u_1\), \(u_2\), ... is defined by $$ u_0=1,\hspace{0.2in} u_1=1, \quad u_{n+1}=u_n+u_{n-1} \hspace{0.2in}{\rm for}\hspace{0.1in}n \ge 1\,. $$ Prove that $$ u^2_{n+2} + u^2_{n-1} = 2( u^2_{n+1} + u^2_n ) \,. $$ Using induction, or otherwise, prove the following result: \[ u_{2n} = u^2_n + u^2_{n-1} \quad \mbox{ and }\quad u_{2n+1} = u^2_{n+1} - u^2_{n-1} \] for any positive integer \(n\).