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Level 2 — RecallCalculus III — Sequences & Series

Calculus III — Sequences & Series

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Level 2 — Recall & Standard Problems

Time limit: 30 minutes
Total marks: 40

Q1. State the definition of convergence of a sequence (an)(a_n) to a limit LL. Then determine whether the sequence an=3n2+12n2+5a_n = \dfrac{3n^2+1}{2n^2+5} converges, and give its limit. (4 marks)

Q2. Use the Squeeze Theorem to evaluate lim⁡n→∞sin⁡nn\displaystyle \lim_{n\to\infty} \frac{\sin n}{n}. State clearly the bounding sequences. (4 marks)

Q3. State the convergence condition for a geometric series ∑n=0∞arn\sum_{n=0}^{\infty} ar^n. Hence find the sum of ∑n=1∞23n.\sum_{n=1}^{\infty} \frac{2}{3^{n}}. (4 marks)

Q4. Evaluate the telescoping series ∑n=1∞1n(n+1)\sum_{n=1}^{\infty} \frac{1}{n(n+1)} by first finding a partial-fraction decomposition and the partial sum SNS_N. (5 marks)

Q5. State the Divergence Test. Use it to show that ∑n=1∞n2n+1\sum_{n=1}^{\infty} \frac{n}{2n+1} diverges. (4 marks)

Q6. State the pp-series result. Determine (with brief justification) whether each of the following converges: (a) ∑1n3/2\displaystyle\sum \frac{1}{n^{3/2}}, (b) ∑1n\displaystyle\sum \frac{1}{\sqrt n}. (4 marks)

Q7. Use the Ratio Test to determine whether ∑n=1∞2nn!\displaystyle\sum_{n=1}^{\infty} \frac{2^n}{n!} converges. (5 marks)

Q8. State the Alternating Series (Leibniz) Test. Use it to show that ∑n=1∞(−1)n+1n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} converges. State whether the convergence is absolute or conditional. (5 marks)

Q9. Write down the Maclaurin series of exe^x and use it to write the first four terms of the series for e−x2e^{-x^2}. (5 marks)

Answer keyMark scheme & solutions

Q1. (4 marks) Definition: (an)→L(a_n)\to L if for every ε>0\varepsilon>0 there exists N∈NN\in\mathbb{N} such that ∣an−L∣<ε|a_n-L|<\varepsilon for all n>Nn>N. (2) Divide numerator and denominator by n2n^2: an=3+1/n22+5/n2→32.a_n=\frac{3+1/n^2}{2+5/n^2}\to \frac{3}{2}. (1) Converges to L=32L=\tfrac{3}{2} (highest powers dominate). (1)

Q2. (4 marks) Since −1≤sin⁡n≤1-1\le \sin n\le 1: −1n≤sin⁡nn≤1n-\dfrac{1}{n}\le \dfrac{\sin n}{n}\le \dfrac{1}{n}. (2) Both bounds →0\to 0 as n→∞n\to\infty. (1) By Squeeze Theorem, lim⁡=0\lim = 0. (1)

Q3. (4 marks) Converges iff ∣r∣<1|r|<1, with sum a1−r\dfrac{a}{1-r}. (2) Here ∑n=1∞2⋅(1/3)n\sum_{n=1}^\infty 2\cdot(1/3)^n: first term a=2/3a=2/3, r=1/3r=1/3. (1) Sum =2/31−1/3=2/32/3=1.=\dfrac{2/3}{1-1/3}=\dfrac{2/3}{2/3}=1. (1)

Q4. (5 marks) 1n(n+1)=1n−1n+1\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}. (2) Partial sum SN=∑n=1N(1n−1n+1)=1−1N+1S_N=\sum_{n=1}^N\left(\frac1n-\frac1{n+1}\right)=1-\dfrac{1}{N+1} (telescoping). (2) lim⁡N→∞SN=1\lim_{N\to\infty}S_N=1, so sum =1=1. (1)

Q5. (4 marks) Divergence Test: if lim⁡n→∞an≠0\lim_{n\to\infty}a_n\ne 0 (or doesn't exist), ∑an\sum a_n diverges. (2) lim⁡n2n+1=12≠0\lim \dfrac{n}{2n+1}=\dfrac12\ne 0. (1) Hence the series diverges. (1)

Q6. (4 marks) pp-series ∑1/np\sum 1/n^p converges iff p>1p>1. (2) (a) p=3/2>1p=3/2>1 → converges. (1) (b) p=1/2≤1p=1/2\le 1 → diverges. (1)

Q7. (5 marks) an=2nn!a_n=\dfrac{2^n}{n!}. Ratio: an+1an=2n+1(n+1)!⋅n!2n=2n+1\dfrac{a_{n+1}}{a_n}=\dfrac{2^{n+1}}{(n+1)!}\cdot\dfrac{n!}{2^n}=\dfrac{2}{n+1}. (3) L=lim⁡n→∞2n+1=0<1L=\lim_{n\to\infty}\dfrac{2}{n+1}=0<1. (1) Since L<1L<1, the series converges (absolutely). (1)

Q8. (5 marks) Leibniz Test: if bn>0b_n>0, bnb_n is decreasing, and bn→0b_n\to 0, then ∑(−1)n+1bn\sum(-1)^{n+1}b_n converges. (2) Here bn=1/nb_n=1/n: positive, decreasing, →0\to 0. (2) So the series converges. Since ∑1/n\sum 1/n (harmonic) diverges, convergence is conditional. (1)

Q9. (5 marks) ex=∑n=0∞xnn!=1+x+x22!+x33!+⋯e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots (2) Substitute x→−x2x\to -x^2: e−x2=∑n=0∞(−x2)nn!=1−x2+x42!−x63!+⋯e^{-x^2}=\sum_{n=0}^\infty \frac{(-x^2)^n}{n!}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots (3) First four terms: 1−x2+x42−x661-x^2+\dfrac{x^4}{2}-\dfrac{x^6}{6}.

[
  {"claim":"Limit of (3n^2+1)/(2n^2+5) is 3/2","code":"n=symbols('n',positive=True); result = limit((3*n**2+1)/(2*n**2+5), n, oo)==Rational(3,2)"},
  {"claim":"Sum of 2/3^n from 1 to inf equals 1","code":"n=symbols('n',positive=True); result = summation(2/3**n,(n,1,oo))==1"},
  {"claim":"Telescoping sum 1/(n(n+1)) from 1 to inf equals 1","code":"n=symbols('n',positive=True); result = summation(1/(n*(n+1)),(n,1,oo))==1"},
  {"claim":"Ratio test limit for 2^n/n! is 0","code":"n=symbols('n',positive=True); result = limit((2/(n+1)),n,oo)==0"},
  {"claim":"Sum of 2^n/n! from 0 to inf equals e^2 (converges)","code":"n=symbols('n',nonnegative=True); result = summation(2**n/factorial(n),(n,0,oo))==exp(2)"}
]