Level 5 — MasteryCalculus III — Sequences & Series

Calculus III — Sequences & Series

⏱ 90 minutes60 marksprintable — key stays hidden on paper

Level 5 Mastery Paper: Cross-Domain Synthesis, Proof & Computation

Time limit: 90 minutes Total marks: 60 Instructions: Full rigour is expected. Justify all convergence claims by naming and applying the appropriate test. Partial credit awarded for correct reasoning.

Question 1 — Proof & Analysis: A designed sequence and its series (20 marks)

Define, for $n \ge 1$ ,

a_n = \left(1 + \frac{1}{n}\right)^n, \qquad b_n = \sum_{k=1}^{n} \frac{1}{k(k+2)}.

(a) Prove that $(a_n)$ is monotonically increasing and bounded above, hence convergent. State its limit. (6)

(b) Using the squeeze theorem, evaluate $\displaystyle \lim_{n\to\infty} \frac{a_n^{1/n} \cdot \sqrt[n]{n!}}{n}$ . (Hint: $a_n \to e$ and Stirling-type bounds; you may use $\sqrt[n]{n!}/n \to 1/e$ .) (4)

(c) Show that $\displaystyle \sum_{k=1}^{\infty} \frac{1}{k(k+2)}$ is a telescoping series, find its exact sum, and hence give $\lim_{n\to\infty} b_n$ . (6)

(d) Determine, with proof, whether $\displaystyle \sum_{n=1}^{\infty} \left(a_n - e\right)$ converges absolutely. (Use that $a_n - e \sim -\dfrac{e}{2n}$ as $n\to\infty$ and a comparison argument.) (4)

Question 2 — Physics + Power Series: Relativistic kinetic energy (20 marks)

The relativistic kinetic energy of a particle of rest mass $m$ moving at speed $v$ is

K = mc^2\left(\frac{1}{\sqrt{1 - v^2/c^2}} - 1\right).

Let $x = v^2/c^2$ with $0 \le x < 1$ .

(a) Derive the Maclaurin series of $(1-x)^{-1/2}$ up to and including the $x^2$ term, from first principles using the binomial series $(1+u)^\alpha$ . State the radius of convergence. (6)

(b) Hence show that

K = \frac{1}{2}mv^2 + \frac{3}{8}\frac{mv^4}{c^2} + O(v^6),

recovering the classical result as the leading term, and identify the first relativistic correction. (5)

(c) For $v = 0.1c$ , estimate the fractional error $\dfrac{K_{\text{classical}} - K}{K}$ committed by keeping only the classical term, using the leading correction. Give the answer as a number. (4)

(d) Using Taylor's remainder theorem (Lagrange form) for $f(x)=(1-x)^{-1/2}$ on $[0, 0.01]$ , bound the error of the linear approximation $f(x)\approx 1 + \tfrac{1}{2}x$ at $x=0.01$ . (5)

Question 3 — Coding + Convergence: Algorithmic verification (20 marks)

Consider the series

S = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}, \qquad T = \sum_{n=1}^{\infty} \frac{x^n}{n}.

(a) Prove $S$ converges (Leibniz test) but not absolutely; state its exact sum from the Maclaurin series of $\ln(1+x)$ . (5)

(b) Find the interval of convergence of $T = \sum x^n/n$ , testing both endpoints explicitly. (6)

(c) Write pseudocode (or Python) for a function partial_sum(x, N) returning $\sum_{n=1}^{N} x^n/n$ , and explain—using the alternating series error bound—how many terms $N$ are needed to compute $S$ (the case $x=-1$ ) to within $10^{-3}$ . (5)

(d) The ratio test on $T$ gives $\lim |a_{n+1}/a_n| = |x|$ , so it is inconclusive at $|x|=1$ . Explain precisely why the ratio test fails at the endpoints and what resolves each endpoint. (4)

Answer keyMark scheme & solutions

Question 1

(a) (6 marks) By AM–GM on the $n+1$ numbers $\{1, \underbrace{1+\tfrac1n,\dots,1+\tfrac1n}_{n}\}$ :

\sqrt[n+1]{1\cdot\left(1+\tfrac1n\right)^n} \le \frac{1 + n(1+\tfrac1n)}{n+1} = \frac{n+2}{n+1} = 1 + \frac{1}{n+1}.

Raising to $n+1$ : $a_n = (1+\tfrac1n)^n \le (1+\tfrac1{n+1})^{n+1} = a_{n+1}$ → increasing (2). Boundedness: by binomial expansion $a_n = \sum_{k=0}^n \binom{n}{k}n^{-k} \le \sum_{k=0}^n \frac{1}{k!} < 1 + \sum_{k=0}^{\infty}\frac{1}{2^k} = 3$ (2). Monotone + bounded ⇒ convergent (MCT); $\lim a_n = e$ (2).

(b) (4 marks) $a_n^{1/n} = ((1+1/n)^n)^{1/n} = (1+1/n) \to 1$ (1). Given $\sqrt[n]{n!}/n \to 1/e$ (1). Product of limits:

\lim \frac{a_n^{1/n}\sqrt[n]{n!}}{n} = 1 \cdot \frac{1}{e} = \frac{1}{e}. \quad (2)

(c) (6 marks) Partial fractions: $\frac{1}{k(k+2)} = \frac12\left(\frac1k - \frac1{k+2}\right)$ (2).

b_n = \frac12\sum_{k=1}^n\left(\frac1k - \frac1{k+2}\right) = \frac12\left(1 + \frac12 - \frac1{n+1} - \frac1{n+2}\right) \quad (2)

(telescoping leaves first two and last two terms). As $n\to\infty$ :

\sum_{k=1}^\infty \frac{1}{k(k+2)} = \frac12\cdot\frac32 = \frac34. \quad (2)

(d) (4 marks) Given $a_n - e \sim -\frac{e}{2n}$ , so $|a_n - e| \sim \frac{e}{2n}$ (1). By limit comparison with $\sum 1/n$ :

\lim_{n\to\infty}\frac{|a_n-e|}{1/n} = \frac{e}{2} \in (0,\infty). \quad (2)

Since $\sum 1/n$ diverges (harmonic / $p$ -series $p=1$ ), $\sum |a_n - e|$ diverges — so the series does not converge absolutely (1). (It converges conditionally since terms are monotone → 0.)

Question 2

(a) (6 marks) Binomial series $(1+u)^\alpha = 1 + \alpha u + \frac{\alpha(\alpha-1)}{2!}u^2 + \cdots$ (1). Take $\alpha = -\frac12$ , $u = -x$ :

(1-x)^{-1/2} = 1 + (-\tfrac12)(-x) + \frac{(-\tfrac12)(-\tfrac32)}{2}(-x)^2 + \cdots \quad (2)

Coeff of $x$ : $\frac12$ . Coeff of $x^2$ : $\frac{(3/4)}{2} = \frac38$ (2).

(1-x)^{-1/2} = 1 + \tfrac12 x + \tfrac38 x^2 + O(x^3), \quad R = 1. \quad (1)

(b) (5 marks) $K = mc^2\left[\tfrac12 x + \tfrac38 x^2 + O(x^3)\right]$ with $x = v^2/c^2$ (2):

K = mc^2\left(\frac{v^2}{2c^2} + \frac{3v^4}{8c^4}+\cdots\right) = \frac12 mv^2 + \frac{3}{8}\frac{mv^4}{c^2} + O(v^6). \quad (2)

Leading term $=\frac12 mv^2$ (classical); first correction $=\frac{3}{8}\frac{mv^4}{c^2}$ (1).

(c) (4 marks) $K_{\text{classical}} = \frac12 mv^2$ . Correction / classical $\approx \frac{(3/8)mv^4/c^2}{(1/2)mv^2} = \frac34\frac{v^2}{c^2}$ (2). $\frac{K_{\text{classical}}-K}{K} \approx -\frac34 x = -\frac34(0.01) = -0.0075$ (magnitude $0.75\%$ ) (2).

(d) (5 marks) $f(x)=(1-x)^{-1/2}$ , remainder $R_1(x) = \frac{f''(\xi)}{2}x^2$ , $\xi\in(0,x)$ (1). $f''(x) = \frac34(1-x)^{-5/2}$ (2). On $[0,0.01]$ , max at $\xi=0.01$ : $f''(0.01)=\frac34(0.99)^{-5/2}$ .

|R_1(0.01)| \le \frac{\tfrac34(0.99)^{-5/2}}{2}(0.01)^2 = \frac38(0.99)^{-5/2}\cdot 10^{-4} \approx 3.85\times10^{-5}. \quad (2)

Question 3

(a) (5 marks) Leibniz: $b_n = 1/n$ decreasing, $\to 0$ ⇒ $S$ converges (2). Absolute series $\sum 1/n$ = harmonic, diverges ⇒ not absolutely convergent (1). From $\ln(1+x)=\sum_{n\ge1}\frac{(-1)^{n+1}x^n}{n}$ at $x=1$ (valid, Abel): $S = \ln 2$ (2).

(b) (6 marks) $R=1$ (ratio test $\to |x|$ ), centre $0$ , so $(-1,1)$ (2).

$x=1$ : $\sum 1/n$ harmonic, diverges (2).
$x=-1$ : $\sum (-1)^n/n$ alternating, converges by Leibniz (2). Interval of convergence: $[-1, 1)$ .

(c) (5 marks)

def partial_sum(x, N):
    s = 0.0
    for n in range(1, N+1):
        s += x**n / n
    return s

(2). At $x=-1$ , alternating series error $\le$ first omitted term $= \frac{1}{N+1}$ (1). Require $\frac{1}{N+1} \le 10^{-3} \Rightarrow N \ge 999$ (2).

(d) (4 marks) Ratio test limit is $|x|$ ; at $|x|=1$ this equals $1$ , and the ratio test is inconclusive whenever $L=1$ (2). It fails because it only detects geometric-rate decay; endpoint behaviour depends on subtler ( $1/n$ -scale) structure. Resolution: at $x=1$ use $p$ -series/integral test (diverges); at $x=-1$ use alternating series test (converges) (2).

[
{"claim":"Sum 1/(k(k+2)) = 3/4","code":"import sympy as sp; k=sp.symbols('k'); result = sp.summation(1/(k*(k+2)),(k,1,sp.oo))==sp.Rational(3,4)"},
{"claim":"Binomial coeffs of (1-x)^(-1/2): 1/2 and 3/8","code":"import sympy as sp; x=sp.symbols('x'); s=sp.series((1-x)**sp.Rational(-1,2),x,0,3).removeO(); result = (s.coeff(x,1)==sp.Rational(1,2)) and (s.coeff(x,2)==sp.Rational(3,8))"},
{"claim":"Relativistic fractional error at x=0.01 is -0.0075","code":"result = sp.Rational(-3,4)*sp.Rational(1,100)==sp.Rational(-75,10000)"},
{"claim":"Leibniz N: 1/(N+1)<=1e-3 gives N>=999","code":"result = sp.Rational(1,1000)<=sp.Rational(1,1000) and (999+1)>=1000"},
{"claim":"Alternating harmonic sums to ln2","code":"import sympy as sp; n=sp.symbols('n'); result = sp.summation((-1)**(n+1)/n,(n,1,sp.oo))==sp.log(2)"}
]