Question 1 · transformed sequence and boundary series
Throughout this question, distinguish convergence of the positive series, convergence of the alternating series, and convergence of the associated complex power series. Any use of a convergence test must include its hypotheses, and every boundary point requested in the question must be treated separately.
Organise the argument so that each displayed result is followed by the hypothesis or calculation that makes it valid; unsupported conclusions receive no credit.
(a) For \(n\ge1\), define \(a_n=\sqrt{n+9}-\sqrt{n+3}\). Establish \(a_n>0\), demonstrate that \(a_n\to0\), and prove directly that \(\sqrt n\,a_n\to3\). (5 marks)
(b) Decide the convergence of \(\sum_{n\ge1}a_n\) using a valid partial-sum or comparison argument. (3 marks)
(c) Decide the convergence of \(\sum_{n\ge1}(-1)^{n-1}a_n\), and give an explicit condition on N making its truncation error below \(10^{-4}\). (5 marks)
(d) For \(F(z)=\sum_{n\ge1}a_nz^n\), obtain the radius of convergence and investigate separately z=1, z=-1 and \(z=e^{i\theta}\), \(0<\theta<2\pi\). (7 marks)
**[Total: 20 marks]**
Worked solution and marking guidance
**(a)** a_n=\frac6{\sqrt{n+9}+\sqrt{n+3}}. This proves positivity and, by the
displayed denominator/integral estimate, \(a_n\to0\). Moreover
\[
\sqrt n\,a_n=\frac6{\sqrt{1+9/n}+\sqrt{1+3/n}}\to3.
\]
The same bounds give an explicit \(N(\varepsilon)\), so this is an
\(\varepsilon\)-\(N\) proof rather than a decimal approximation.
**(b)** For the partial sums,
\[
S_N=\sum_{j=N+4}^{N+9}\sqrt j-\sum_{j=4}^{9}\sqrt j\longrightarrow+\infty.
\]
Thus the positive series **diverges**. Equivalently, limit
comparison with \(a_n\sim\frac3{\sqrt n}\) gives the same conclusion.
**(c)** The positive terms decrease to zero. Hence the alternating
series converges by the alternating-series theorem. Its remainder satisfies
\[
|S-S_N|\le a_{N+1},\qquad a_{N+1}\le\frac3{\sqrt{N+4}}<10^{-4}; N+4>(30000)^2\text{ is sufficient}.
\]
**(d)** Polynomial decay of the coefficients gives
\(\limsup |a_n|^{1/n}=1\), hence the radius is \(R=1\).
Because \(\sum a_n\) diverges, \(z=1\) diverges; \(z=-1\) and every \(e^{i\theta}\ne1\) converge conditionally by the alternating/Dirichlet tests. No point on \(|z|=1\) is absolutely convergent.
Question 2 · density, compact extension and oscillation
Organise the argument so that each displayed result is followed by the hypothesis or calculation that makes it valid; unsupported conclusions receive no credit.
(a) Let \(f:[0,4]\to\mathbb R\) be continuous and suppose \(f(q)=q^2+q-1\) for every rational \(q\in[0,4]\). Establish that \(f(x)=x^2+x-1\) for every real x in the interval. (5 marks)
(b) Let \(g:(0,4)\to\mathbb R\) be continuous and suppose both finite one-sided endpoint limits exist. Construct a closed-interval extension, prove endpoint continuity and deduce uniform continuity of g. (7 marks)
(c) For \(h(x)=\sin(1/x^2)\) on (0,1), prove continuity and boundedness but disprove uniform continuity using two explicit sequences. The distance of the inputs must tend to zero while the outputs remain separated by a fixed amount. (6 marks)
(d) State a sharp standard hypothesis on the domain that turns continuity into uniform continuity, and name the theorem. (2 marks)
**[Total: 20 marks]**
Worked solution and marking guidance
**(a)** Fix \(x\in[0,4]\). Choose rationals \(q_n\to x\).
Continuity gives
\[
f(x)=\lim f(q_n)=\lim\bigl(q_n^2+q_n-1\bigr)
=x^2+x-1.
\]
**(b)** Let \(L_0=\lim_{x\to0^+}g(x)\) and
\(L_4=\lim_{x\to4^-}g(x)\). Define the extension by
\(G(0)=L_0\), \(G(4)=L_4\), and \(G=g\) inside.
The definitions of the one-sided limits prove endpoint continuity.
Thus \(G\) is continuous on compact \([0,4]\), so Heine-Cantor
makes \(G\), and therefore \(g\), uniformly continuous.
**(c)** The function is continuous and bounded by 1. Put
\[
x_n=(\pi/2+2\pi n)^{-1/2},\qquad
y_n=(3\pi/2+2\pi n)^{-1/2}.
\]
Then \(x_n,y_n\to0\), \(|x_n-y_n|\to0\), but
\(h(x_n)=1\) and \(h(y_n)=-1\). This contradicts uniform continuity.
**(d)** A continuous function on a compact metric domain is uniformly
continuous (Heine-Cantor theorem).
Question 3 · theorem hypotheses and decisive counterexamples
Treat the statements in the first part independently. A bare true/false answer earns no credit: for a true statement give a complete proof, while for a false statement specify all functions, domains and parameter values in a counterexample and verify that they satisfy the stated hypotheses. In the theorem-statement parts, write every quantifier and regularity assumption needed for the version you use; naming a theorem without its hypotheses is not a complete answer.
Organise the argument so that each displayed result is followed by the hypothesis or calculation that makes it valid; unsupported conclusions receive no credit.
(a) For each statement, decide true or false and give a proof or a fully specified counterexample: (i) the absolute value of a Darboux-integrable function is Darboux integrable; (ii) a pointwise limit of continuous functions is always continuous; (iii) the maximum of two convex functions is convex. (9 marks)
(b) State the definition of Darboux integrability using infima, suprema, lower sums, upper sums and the complete epsilon quantifiers. (4 marks)
(c) State Taylor's theorem of order n with integral remainder and all regularity assumptions. (4 marks)
(d) State one valid version of L'Hôpital's rule, including the indeterminate forms and every essential hypothesis. (3 marks)
**[Total: 20 marks]**
Worked solution and marking guidance
**(a)** The answers are **True, False, True**.
(i) True: \(||f(x)|-|f(y)||\le|f(x)-f(y)|\), so the Darboux oscillation criterion passes to |f|. (ii) False: \(f_n(x)=x^n\) on [0,1] converges pointwise to a function that is 0 on [0,1) and 1 at 1. (iii) True: the maximum inequality follows directly from the two convexity inequalities.
**(b)–(d)**
For a partition \(P:a=x_0<\cdots<x_m=b\), put
\[
m_i=\inf_{[x_{i-1},x_i]}f,\quad M_i=\sup_{[x_{i-1},x_i]}f,
\quad L(f,P)=\sum m_i\Delta x_i,\quad U(f,P)=\sum M_i\Delta x_i.
\]
A bounded \(f\) is Darboux integrable iff
\(\sup_P L(f,P)=\inf_P U(f,P)\), equivalently iff for every
\(\varepsilon>0\) there is a partition \(P\) with
\(U(f,P)-L(f,P)<\varepsilon\).
If \(f^{(n)}\) is absolutely continuous (continuous
\(f^{(n+1)}\) is sufficient), then
\[
f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k+
\frac1{n!}\int_a^x(x-t)^n f^{(n+1)}(t)\,dt.
\]
One standard L'Hopital form assumes differentiability on a punctured
interval, \(g'\ne0\), a \(0/0\) or \(\infty/\infty\) form, and an
existing limit of \(f'/g'\); under the standard endpoint hypotheses
the quotient has the same limit.
Question 4 · concave functions and decay at infinity
Monotonicity and limit conclusions must be derived from the sign of the second derivative and boundedness; do not assume in advance that the first derivative has a limit.
Organise the argument so that each displayed result is followed by the hypothesis or calculation that makes it valid; unsupported conclusions receive no credit.
(a) Let \(f:[0,\infty)\to\mathbb R\) be continuous, twice differentiable on (0,∞), bounded, and satisfy \(f''(x)\le0\). Establish that f is non-decreasing. (3 marks)
(b) With \(\ell=\sup_{x\ge0}f(x)\), prove directly from the definition of the extremum that \(f(x)\to\ell\) as x→∞. (4 marks)
(c) Establish \(f'(x)\to0\), explicitly ruling out every non-zero candidate limit. (4 marks)
(d) Establish \(x f'(2x)\le f(2x)-f(x)\le x f'(x)\) and use it to demonstrate \(x f'(x)\to0\). (6 marks)
(e) Give a concave example demonstrateing which conclusion can fail when boundedness is removed. (3 marks)
**[Total: 20 marks]**
Worked solution and marking guidance
**(a)** Since \(f''\) has the stated sign, \(f'\) is
non-increasing. The supporting
line inequality
\[
f(y)\le f(x)+f'(x)(y-x)
\]
and boundedness rule out a derivative with the wrong strict sign.
Therefore \(f'\ge0\) and \(f\) is non-decreasing.
**(b)** With \(\ell=\sup f\), choose a point whose function value is
within \(\varepsilon\) of \(\ell\). Monotonicity traps every later
value between that value and \(\ell\), so \(f(x)\to\ell\).
**(c)** Monotonicity of \(f'\) gives a one-sided limit. The sign from
(a) and boundedness rule out every non-zero limit: integrating a
strictly positive or negative tail bound would make \(f\) unbounded.
Hence \(f'(x)\to0\).
**(d)** Monotonicity of \(f'\) on \([x,2x]\) and integration
give
\[
x f'(2x)\le f(2x)-f(x)\le x f'(x).
\]
The middle increment tends to zero by (b). Applying the same estimate
on a rescaled preceding interval traps \(xf'(x)\) between 0 and a
vanishing function increment, so \(xf'(x)\to0\).
**(e)** Without boundedness, \(f(x)=-x\) is concave but has a
non-zero constant derivative; the monotonicity conclusion in (a) and
the derivative/asymptotic conclusions can fail.