(i)     Show that the improper integral \(\int_0^\infty  {\frac{1}{{{x^2} + 1}}} {\rm{d}}x\) is convergent.

(ii)     Use the integral test to deduce that the series \(\sum\limits_{n = 0}^\infty  {\frac{1}{{{n^2} + 1}}} \) is convergent, giving reasons why this test can be applied.

(i)     Show that the series \(\sum\limits_{n = 0}^\infty  {\frac{{{{( - 1)}^n}}}{{{n^2} + 1}}} \) is convergent.

(ii)     If the sum of the above series is \(S\), show that \(\frac{3}{5} < S < \frac{2}{3}\) .

For the series \(\sum\limits_{n = 0}^\infty  {\frac{{{x^n}}}{{{n^2} + 1}}} \)

  (i)     determine the radius of convergence;

  (ii)     determine the interval of convergence using your answers to (b) and (c).

(i)     consider \(\int_0^R {\frac{1}{{{x^2} + 1}}} {\rm{d}}x\)     M1 

\( = \left[ {\arctan (x)} \right]_0^R = \arctan (R)\)     A1

\(\mathop {\lim }\limits_{R \to \infty } \arctan (R) = \frac{\pi }{2}\) (a finite number)     R1

hence the improper integral is convergent     AG

(ii)     the terms of the series are positive     A1

the terms are decreasing    A1

the terms tend to zero     A1

by the integral test, the series converges     AG

[6 marks]

(i)     the absolute values of the terms are monotonically decreasing     A1

to zero     A1

the series converges by the alternating series test     R1AG

Note: Accept absolute convergence, with reference to part (b)(ii) \( \Rightarrow \) convergence.

(ii)     statement that successive partial sums bound the total sum     R1

\(S > \frac{1}{1} - \frac{1}{2} + \frac{1}{5} - \frac{1}{{10}} = \frac{3}{5}\)     A1

\(S < \frac{1}{1} - \frac{1}{2} + \frac{1}{5} - \frac{1}{{10}} + \frac{1}{{17}} = 0.6588\)     A1

\(S < 0.6588 < \frac{2}{3}\)     AG

[6 marks]

(i)     consider \(\left| {\frac{{\frac{{{x^{n + 1}}}}{{{{(n + 1)}^2} + 1}}}}{{\frac{{{x^n}}}{{{n^2} + 1}}}}} \right|\)     M1 

\( = \left| {\frac{{x({n^2} + 1)}}{{{{(n + 1)}^2} + 1}}} \right|\)     A1  

\( \to \left| x \right|\) as \(n \to \infty \)     A1

therefore radius of convergence \( = 1\)     A1

(ii)     interval of convergence \( = \left[ { - 1,1} \right]\)     A1A1

Note: A1 for [\( - 1\), and A1 for \(1\)].

[6 marks]

Although the various parts of this question were algebraically uncomplicated, many candidates revealed their lack of understanding of the necessary rigour required in the analysis of limits, improper integrals and the testing of series for convergence. In (b)(i), the upper limit in the integral was often taken as infinity, without any mention of an underlying limiting process.

Although the various parts of this question were algebraically uncomplicated, many candidates revealed their lack of understanding of the necessary rigour required in the analysis of limits, improper integrals and the testing of series for convergence.

Although the various parts of this question were algebraically uncomplicated, many candidates revealed their lack of understanding of the necessary rigour required in the analysis of limits, improper integrals and the testing of series for convergence. Many candidates were more confident with part (d) than with the other parts of the question.