Find the set of values of x for which the series \(\sum\limits_{n = 1}^\infty  {{{\left( {\frac{{2x}}{{x + 1}}} \right)}^n}} \) has a finite sum.

Hence find the sum in terms of x.

for the series to have a finite sum, \(\left| {\frac{{2x}}{{x + 1}}} \right| < 1\)     R1

(sketch from gdc or algebraic method)     M1

\({S_\infty }\) exists when \( - \frac{1}{3} < x < 1\)     A1A1

Note: Award A1 for bounds and A1 for strict inequalities.

[4 marks]

\({S_\infty } = \frac{{\frac{{2x}}{{x + 1}}}}{{1 - \frac{{2x}}{{x + 1}}}} = \frac{{2x}}{{1 - x}}\)     M1A1

[2 marks]

A large number of candidates omitted the absolute value sign in the inequality in (a), or the use of the correct double inequality. Among candidates who had the correct statement, those who used their GDC were the most successful. The algebraic solution of the inequality was difficult for some candidates. In (b), quite a number of candidates found the sum of the first n terms of the geometric series, rather than the infinite sum of the series.

A large number of candidates omitted the absolute value sign in the inequality in (a), or the use of the correct double inequality. Among candidates who had the correct statement, those who used their GDC were the most successful. The algebraic solution of the inequality was difficult for some candidates. In (b), quite a number of candidates found the sum of the first n terms of the geometric series, rather than the infinite sum of the series.