IB DP Further Mathematics 5.2 HL Paper 1

 

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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)     \(\frac{\pi }{6}\) .

[11]
b.

Show that \(\sum\limits_{n = 1}^{100} {n! \equiv 3(\bmod 15)} \) .

[4]
c.
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]

b.

\(\sum\limits_{n = 1}^{100} {n! = 1! + 2! + 3! + 4! + 5! +  \ldots } \)     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]

c.

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.

[4]
a.

State the converse proposition.

[1]
b.i.

By giving a suitable example, show that it is false.

[1]
b.ii.
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]

a.

the converse proposition is that if \(\sum {u_n^2} \) is convergent, then \(\sum {{u_n}} \) is also convergent     A1

[1 mark]

b.i.

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]

b.ii.

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.

[6]
a.

Show that \(S > 0.4\) .

[2]
b.
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]

a.

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]

b.
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