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Question
(i) Sum the series \(\sum\limits_{r = 0}^\infty {{x^r}} \) .
(ii) Hence, using sigma notation, deduce a series for
(a) \(\frac{1}{{1 + {x^2}}}\) ;
(b) \(\arctan x\) ;
(c) \(\frac{\pi }{6}\) .
Show that \(\sum\limits_{n = 1}^{100} {n! \equiv 3(\bmod 15)} \) .
Answer/Explanation
Markscheme
(i) \(\sum\limits_{r = 0}^\infty {{x^r}} = 1 + x + {x^2} + {x^3} + {x^4} + \ldots = \frac{1}{{1 – x}}\) A1
(ii) (a) replacing x by \( – {x^2}\) gives (M1)
\(\frac{1}{{1 – ( – {x^2})}} = 1 + ( – {x^2}) + {( – {x^2})^2} + {( – {x^2})^3} + {( – {x^2})^4} + \ldots \) A1
\(\frac{1}{{1 + {x^2}}} = 1 – {x^2} + {x^4} – {x^6} + {x^8} – \ldots \) (A1)
\( = \sum\limits_{r = 0}^\infty {{{( – 1)}^r}{x^{2r}}} \) A1 N2
(b) \(\arctan x = \int {\frac{{{\rm{d}}x}}{{1 + {x^2}}}} = x – \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} – \frac{{{x^7}}}{7} + \ldots + c\) M1A1
\(x = 0 \Rightarrow c = 0\) A1
\(\arctan x = \sum\limits_{r = 0}^\infty {{{( – 1)}^r}\frac{{{x^{2r + 1}}}}{{2r + 1}}} \) A1
(c) by taking \(x = \frac{1}{{\sqrt 3 }}\) M1
\(\arctan \frac{1}{{\sqrt 3 }} = \frac{\pi }{6} = \sum\limits_{r = 0}^\infty {\frac{{{{( – 1)}^r}{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^{2r + 1}}}}{{2r + 1}}} \) A1
[11 marks]
\(\sum\limits_{n = 1}^{100} {n! = 1! + 2! + 3! + 4! + 5! + \ldots } \) M1
\( = 1 + 2 + 6 + 24 + 120 + \ldots \)
\( \equiv 1 + 2 + 6 + 24 + 0 + 0 + 0 + \ldots (\bmod 15)\) M1A1
\( \equiv 33(\bmod 15)\) A1
\( \equiv 3(\bmod 15)\) AG
[4 marks]
Question
Consider the infinite series \(S = \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{{2^{2n}}\left( {2{n^2} – 1} \right)}}} \).
(a) Determine the radius of convergence.
(b) Determine the interval of convergence.
Answer/Explanation
Markscheme
(a) let \({T_n}\) denote the \(n{\text{th}}\) term
consider
\(\frac{{{T_{n + 1}}}}{{{T_n}}} = \frac{{{x^{(n + 1)}}}}{{{2^{2(n + 1)}}\left( {2{{[n + 1]}^2} – 1} \right)}} \times \frac{{{2^2}\left( {2{n^2} – 1} \right)}}{{{x^n}}}\) M1
\( = \frac{x}{{{2^2}}} \times \frac{{\left( {2{n^2} – 1} \right)}}{{\left( {2{{[n + 1]}^2} – 1} \right)}}\) A1
\( \to \frac{x}{4}\) as \(n \to \infty \) A1
so the radius of convergence is \(4\) A1
[4 marks]
(b) we need to consider \(x = \pm 4\) R1
\(S(4) = \sum\limits_{n = 1}^\infty {\frac{1}{{\left( {2{n^2} – 1} \right)}}} \) A1
\(S(4) < \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) M1
\(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) is convergent; therefore by the comparison test \(S(4)\) is convergent R1
\(S( – 4) = \sum\limits_{n = 1}^\infty {\frac{{{{( – 1)}^n}}}{{\left( {2{n^2} – 1} \right)}}} \) A1
EITHER
this series is convergent because it is absolutely convergent R1
OR
this series is alternating and is convergent R1
THEN
the interval of convergence is therefore \(\left[ { – 4,{\text{ }}4} \right]\) A1
Note: The final A1 is independent of any of the previous marks.
[7 marks]
Question
Find the interval of convergence of the series \(\sum\limits_{k = 1}^\infty {\frac{{{{(x – 3)}^k}}}{{{k^2}}}} \).
Answer/Explanation
Markscheme
ratio test \(\mathop {\lim }\limits_{k \to \infty } \left| {\frac{{{u_{k + 1}}}}{{{u_k}}}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{{{(x – 3)}^{k + 1}}{k^2}}}{{{{(k + 1)}^2}{{(x – 3)}^k}}}} \right|\) M1A1
\(\mathop {\lim }\limits_{k \to \infty } \left| {(x – 3)\frac{{{k^2}}}{{{{(k + 1)}^2}}}} \right|\)
Note: Condone absence of limits and modulus signs in above.
\(\left| {x – 3} \right|\mathop {\lim }\limits_{k \to \infty } \left| {{{\left( {\frac{1}{{1 + \frac{1}{k}}}} \right)}^2}} \right| – \left| {x – 3} \right|\) A1
for convergence \(\left| {x – 3} \right| < 1\) (M1)
\( \Rightarrow – 1 < x – 3 < 1\)
\( \Rightarrow 2 < x < 4\) (A1)
now we need to test end points. (M1)
when \(x = 4\) we have \(\sum\limits_{k = 1}^\infty {\frac{1}{{{k^2}}}} \) which is a convergent series R1
when \(x = 2\) we have \(\sum\limits_{k = 1}^\infty {\frac{{{{( – 1)}^k}}}{{{k^2}}} = – 1 + \frac{1}{{{2^2}}} – \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \ldots } \) which is convergent R1
(alternating series/absolutely converging series)
hence the interval of convergence is \([2,{\text{ }}4]\) A1
Question
Given that the series \(\sum\limits_{n = 1}^\infty {{u_n}} \) is convergent, where \({u_n} > 0\), show that the series \(\sum\limits_{n = 1}^\infty {u_n^2} \) is also convergent.
State the converse proposition.
By giving a suitable example, show that it is false.
Answer/Explanation
Markscheme
since \(\sum {{u_n}} \) is convergent, it follows that \(\mathop {\lim }\limits_{n \to \infty } {u_n} = 0\) R1
therefore, there exists \(N\) such that for \(n \geqslant N,{\text{ }}{u_n} < 1\) R1
Note: Accept as \(n\) gets larger, eventually \({u_n} < 1\).
therefore (for \(n \geqslant N\)), \(u_n^2 < {u_n}\) R1
by the comparison test, \(\sum {u_n^2} \) is convergent R1
[4 marks]
the converse proposition is that if \(\sum {u_n^2} \) is convergent, then \(\sum {{u_n}} \) is also convergent A1
[1 mark]
a suitable counter-example is
\({u_n} = \frac{1}{n}{\text{ }}\left( {{\text{for which }}\sum {u_n^2} {\text{ is convergent but }}\sum {{u_n}} {\text{ is not convergent}}} \right)\) A1
[1 mark]
Question
Use the integral test to determine whether or not \(\sum\limits_{n = 2}^\infty {\frac{1}{{n{{\left( {{\text{ln}}\,n} \right)}^2}}}} \) converges.
Answer/Explanation
Markscheme
let \(u = {\text{ln}}\,x\) (M1)
\( \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x}\)
\(\int {\frac{1}{{x{{\left( {{\text{ln}}\,x} \right)}^2}}}} \,{\text{d}}x = \int {\frac{1}{{{u^2}}}} \,{\text{d}}u\) (A1)
\( = – \frac{1}{u} = – \frac{1}{{{\text{ln}}\,x}}\) (A1)
\(\int\limits_2^m {\frac{1}{{x{{\left( {{\text{ln}}\,x} \right)}^2}}}} \,{\text{d}}x = \left[ { – \frac{1}{{{\text{ln}}\,x}}} \right]_2^m\) M1
\( = \left[ { – \frac{1}{{{\text{ln}}\,m}} + – \frac{1}{{{\text{ln}}\,2}}} \right]\) A1
as \(m \to \infty \), \( – \frac{1}{{{\text{ln}}\,m}} \to 0\) (A1)
\(\int\limits_2^\infty {\frac{1}{{x{{\left( {{\text{ln}}\,x} \right)}^2}}}} \,{\text{d}}x = \frac{1}{{{\text{ln}}\,2}}\) and hence the series converges R1
[7 marks]
Question
Consider the infinite series \(S = \sum\limits_{n = 1}^\infty {{{( – 1)}^{n + 1}}\sin } \left( {\frac{1}{n}} \right)\) .
Show that the series is conditionally convergent but not absolutely convergent.
Show that \(S > 0.4\) .
Answer/Explanation
Markscheme
\(\sin \left( {\frac{1}{n}} \right)\) decreases as n increases A1
\(\sin \left( {\frac{1}{n}} \right) \to 0\) as \(n \to \infty \) A1
so using the alternating series test, the series is conditionally convergent R1
comparing (the absolute series) with the harmonic series:
\(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \) M1
\(\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \left( {\frac{1}{n}} \right)}}{{\frac{1}{n}}} = 1\) A1
since the harmonic series is divergent, it follows by the limit comparison theorem that the given series is not absolutely convergent R1
hence the series is conditionally convergent AG
[6 marks]
successive partial sums are
\(0.841…\)
\(0.362…\)
\(0.689…\)
\(0.441…\) A1
since \(S\) lies between any pair of successive partial sums, it follows that \(S\) lies between \(0.441…\) and \(0.689…\) R1
and is therefore greater than \(0.4\) AG
Note: Use of the facts that the error is always less than the modulus of the next term, or the sequence of even partial sums gives lower bounds are equally acceptable.
[2 marks]