Level 4 — ApplicationCalculus III — Sequences & Series

Calculus III — Sequences & Series

50 marksprintable — key stays hidden on paper

Level 4 Examination (Application)

Time: 60 minutes
Total marks: 50
Instructions: Attempt all questions. Full justification required; state every test/theorem you invoke. No hints are given.

Question 1. (10 marks)

Consider the series $\sum_{n=1}^{\infty}\frac{(-1)^{n}\,n}{n^{2}+1}.$

(a) Determine whether the series converges absolutely, converges conditionally, or diverges. Justify each claim with a named test. (7)

(b) The alternating series test gives an error bound for the truncation error. Estimate how many terms are needed so that the partial sum approximates the series' value with error less than $0.05$ . (3)

Question 2. (12 marks)

(a) Find the radius of convergence and the full interval of convergence of $\sum_{n=1}^{\infty}\frac{(x-2)^{n}}{n\,3^{n}}.$ You must test both endpoints explicitly. (8)

(b) Let $f(x)$ denote the sum of the series in (a) on its interval of convergence. By differentiating term-by-term, express $f'(x)$ as a closed-form elementary function, stating where the identity is valid. (4)

Question 3. (10 marks)

Evaluate the limit $L=\lim_{x\to 0}\frac{e^{x^{2}}-\cos x-\tfrac{3}{2}x^{2}}{x^{4}}$ using Maclaurin series. Show the series expansions to sufficient order and justify why higher-order terms may be neglected. (10)

Question 4. (10 marks)

Define the sequence $(a_n)$ by $a_1=2,\qquad a_{n+1}=\frac{1}{2}\left(a_n+\frac{3}{a_n}\right),\quad n\ge 1.$

(a) Show that $a_n\ge\sqrt{3}$ for all $n\ge1$ , and that $(a_n)$ is monotonically decreasing for $n\ge1$ . (6)

(b) Deduce that $(a_n)$ converges and find its limit. (4)

Question 5. (8 marks)

Use Taylor's remainder theorem to approximate $\ln(1.2)$ using the Maclaurin polynomial of $\ln(1+x)$ of degree $3$ . Then bound the error using the Lagrange remainder, and confirm your numerical bound is consistent with the true error. (8)

Answer keyMark scheme & solutions

Question 1 (10)

(a) Terms: $a_n=\dfrac{(-1)^n n}{n^2+1}$ , with $b_n=\dfrac{n}{n^2+1}>0$ .

Absolute convergence? Consider $\sum \dfrac{n}{n^2+1}$ . Limit comparison with $\sum \frac1n$ : $\lim_{n\to\infty}\frac{n/(n^2+1)}{1/n}=\lim\frac{n^2}{n^2+1}=1\in(0,\infty).$ Since $\sum \frac1n$ diverges (harmonic / $p$ -series $p=1$ ), the series does not converge absolutely. (3)

Conditional convergence — Leibniz test. Need $b_n\to0$ and $b_n$ eventually decreasing.

$b_n=\frac{n}{n^2+1}\to0$ . (1)
Let $g(x)=\frac{x}{x^2+1}$ , $g'(x)=\frac{1-x^2}{(x^2+1)^2}<0$ for $x>1$ , so $b_n$ decreasing for $n\ge1$ . (2)

Hence by the alternating series test the series converges conditionally. (1)

(b) Leibniz error bound: $|S-S_N|\le b_{N+1}=\dfrac{N+1}{(N+1)^2+1}$ . Require $\frac{N+1}{(N+1)^2+1}<0.05$ . For $N=18$ : $b_{19}=19/362\approx0.0525$ (too big). For $N=19$ : $b_{20}=20/401\approx0.0499<0.05$ . ✓ So 20 terms ( $N=19$ , using $S_{19}$ approximation up to term $n=20$ 's bound — i.e. summing through $n=19$ ensures next term $b_{20}<0.05$ ). Accept $N=19$ terms giving bound $b_{20}$ . (3)

Question 2 (12)

(a) Centre $x=2$ , coefficients $c_n=\frac{1}{n3^n}$ . Ratio test on $u_n=\frac{(x-2)^n}{n3^n}$ : $\lim\left|\frac{u_{n+1}}{u_n}\right|=\lim\frac{n}{n+1}\cdot\frac{|x-2|}{3}=\frac{|x-2|}{3}.$ Converges when $|x-2|<3\Rightarrow R=3$ . (4)

Interval base: $|x-2|<3\Rightarrow -1<x<5$ .

Endpoint $x=5$ : $\sum\frac{3^n}{n3^n}=\sum\frac1n$ diverges (harmonic). (2)

Endpoint $x=-1$ : $\sum\frac{(-3)^n}{n3^n}=\sum\frac{(-1)^n}{n}$ converges (alternating harmonic). (1)

Interval of convergence: $\boxed{[-1,\,5)}$ . (1)

(b) Differentiate term-by-term (valid on open interval $(-1,5)$ ): $f'(x)=\sum_{n=1}^\infty\frac{n(x-2)^{n-1}}{n3^n}=\sum_{n=1}^\infty\frac{(x-2)^{n-1}}{3^n}=\frac13\sum_{k=0}^\infty\left(\frac{x-2}{3}\right)^{k}.$ Geometric with ratio $\frac{x-2}{3}$ , $|x-2|<3$ : $f'(x)=\frac13\cdot\frac{1}{1-\frac{x-2}{3}}=\frac{1}{3-(x-2)}=\frac{1}{5-x},\quad -1<x<5.$ (4)

Question 3 (10)

Expansions: $e^{x^2}=1+x^2+\frac{x^4}{2}+O(x^6),\qquad \cos x=1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6).$ (4)

Numerator: $e^{x^2}-\cos x-\tfrac32x^2 = \Big(1+x^2+\tfrac{x^4}{2}\Big)-\Big(1-\tfrac{x^2}{2}+\tfrac{x^4}{24}\Big)-\tfrac32x^2+O(x^6).$ Constant: $1-1=0$ . $x^2$ : $x^2+\tfrac12x^2-\tfrac32x^2=0$ . $x^4$ : $\tfrac12-\tfrac1{24}=\tfrac{12-1}{24}=\tfrac{11}{24}$ . (4)

So numerator $=\frac{11}{24}x^4+O(x^6)$ , hence $L=\lim_{x\to0}\frac{\frac{11}{24}x^4+O(x^6)}{x^4}=\frac{11}{24}.$ Higher-order terms are $O(x^6)/x^4=O(x^2)\to0$ , so negligible. (2)

$\boxed{L=\frac{11}{24}}$

Question 4 (10)

(a) By AM–GM (or algebra): for $a_n>0$ , $a_{n+1}=\tfrac12\!\left(a_n+\tfrac3{a_n}\right)\ge\sqrt{a_n\cdot\tfrac3{a_n}}=\sqrt3.$ Since $a_1=2>0$ , induction gives $a_n>0$ hence $a_{n+1}\ge\sqrt3$ ; also $a_1=2>\sqrt3$ . So $a_n\ge\sqrt3$ all $n$ . (3)

Monotone decreasing: for $a_n\ge\sqrt3$ , $a_n-a_{n+1}=a_n-\tfrac12 a_n-\tfrac{3}{2a_n}=\frac{a_n}{2}-\frac{3}{2a_n}=\frac{a_n^2-3}{2a_n}\ge0.$ Since $a_n^2\ge3$ , difference $\ge0$ , so $(a_n)$ decreasing. (3)

(b) $(a_n)$ is decreasing and bounded below by $\sqrt3$ ⇒ converges (monotone convergence theorem) to limit $L\ge\sqrt3$ . (2) Taking limits: $L=\tfrac12(L+3/L)\Rightarrow 2L^2=L^2+3\Rightarrow L^2=3\Rightarrow L=\sqrt3$ . (2)

Question 5 (8)

Maclaurin: $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$ Degree-3 poly at $x=0.2$ : $P_3(0.2)=0.2-\frac{0.04}{2}+\frac{0.008}{3}=0.2-0.02+0.0026\overline{6}=0.182666\ldots$ (3)

Lagrange remainder: $R_3=\frac{f^{(4)}(\xi)}{4!}x^4$ , $f^{(4)}(x)=-\frac{6}{(1+x)^4}$ , $\xi\in(0,0.2)$ . $|R_3|=\frac{6}{(1+\xi)^4}\cdot\frac{(0.2)^4}{24}=\frac{(0.2)^4}{4(1+\xi)^4}\le\frac{0.0016}{4}=0.0004.$ (3)

True value $\ln1.2=0.1823215\ldots$ ; actual error $=|0.1823215-0.1826667|=0.000345<0.0004$ . ✓ Consistent. (2)

[
  {"claim":"Q2 sum f'(x)=1/(5-x) via geometric series", "code":"x=symbols('x'); s=Rational(1,3)*summation((( x-2)/3)**k,(k:=symbols('k',nonnegative=True),0,oo)); result = simplify(s - 1/(5-x))==0"},
  {"claim":"Q3 limit equals 11/24", "code":"x=symbols('x'); L=limit((exp(x**2)-cos(x)-Rational(3,2)*x**2)/x**4, x, 0); result = (L==Rational(11,24))"},
  {"claim":"Q4 limit satisfies L=sqrt(3)", "code":"L=symbols('L',positive=True); sol=solve(Eq(L, Rational(1,2)*(L+3/L)), L); result = (sqrt(3) in sol)"},
  {"claim":"Q5 P3(0.2) approx 0.182667", "code":"val=Rational(2,10)-Rational(2,10)**2/2+Rational(2,10)**3/3; result = abs(float(val)-0.1826667)<1e-5"},
  {"claim":"Q5 error bound 0.0004 exceeds true error", "code":"result = abs(0.1826667 - float(log(Rational(12,10)))) < 0.0004"}
]