Using the integral test, show that $\sum\limits_{n = 1}^\infty  {\frac{1}{{4{n^2} + 1}}}$ is convergent.

(i)     Show, by means of a diagram, that $\sum\limits_{n = 1}^\infty  {\frac{1}{{4{n^2} + 1}}}  < \frac{1}{{4 \times {1^2} + 1}} + \int_1^\infty  {\frac{1}{{4{x^2} + 1}}{\text{d}}x}$.

(ii)     Hence find an upper bound for $\sum\limits_{n = 1}^\infty  {\frac{1}{{4{n^2} + 1}}}$

$\int {\frac{1}{{4{x^2} + 1}}{\text{d}}x = \frac{1}{2}\arctan 2x + k}$     (M1)(A1)

Note: Do not penalize the absence of “+k”.

$\int_1^\infty  {\frac{1}{{4{x^2} + 1}}{\text{d}}x = \frac{1}{2}\mathop {\lim }\limits_{a \to \infty } } [\arctan 2x]_1^a$     (M1)

Note: Accept $\frac{1}{2}[\arctan 2x]_1^\infty$.

$= \frac{1}{2}\left( {\frac{\pi }{2} - \arctan 2} \right)\,\,\,\,\,( = 0.232)$     A1

hence the series converges     AG

[4 marks]

(i)

     A2

The shaded rectangles lie within the area below the graph so that $\sum\limits_{n = 2}^\infty  {\frac{1}{{4{n^2} + 1}}}  < \int_1^\infty  {\frac{1}{{4{x^2} + 1}}{\text{d}}x}$. Adding the first term in the series, $\frac{1}{{4 \times {1^2} + 1}}$, gives $\sum\limits_{n = 1}^\infty  {\frac{1}{{4{n^2} + 1}}}  < \frac{1}{{4 \times {1^2} + 1}} + \int_1^\infty  {\frac{1}{{4{x^2} + 1}}{\text{d}}x}$     R1AG

(ii)     upper bound $= \frac{1}{5} + \frac{1}{2}\left( {\frac{\pi }{2} - \arctan 2} \right)\,\,\,\,\,( = 0.432)$     A1

[4 marks]

This proved to be a hard question for most candidates. A number of fully correct answers to (a) were seen, but a significant number were unable to integrate ${\frac{1}{{4{x^2} + 1}}}$ successfully. Part (b) was found the hardest by candidates with most candidates unable to draw a relevant diagram, without which the proof of the inequality was virtually impossible.

This proved to be a hard question for most candidates. A number of fully correct answers to (a) were seen, but a significant number were unable to integrate ${\frac{1}{{4{x^2} + 1}}}$ successfully. Part (b) was found the hardest by candidates with most candidates unable to draw a relevant diagram, without which the proof of the inequality was virtually impossible.