Find the radius of convergence of the series \(\sum\limits_{n = 0}^\infty  {\frac{{{{( - 1)}^n}{x^n}}}{{(n + 1){3^n}}}} \).

Determine whether the series \(\sum\limits_{n = 0}^\infty  {\left( {\sqrt[3]{{{n^3} + 1}} - n} \right)} \) is convergent or divergent.

The ratio test gives

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{( - 1)}^{n + 1}}{x^{n + 1}}(n + 1){3^n}}}{{(n + 2){3^{n + 1}}{{( - 1)}^n}{x^n}}}} \right|\)     M1A1

\( = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{(n + 1)x}}{{3(n + 2)}}} \right|\)     A1

\( = \frac{{\left| x \right|}}{3}\)     A1

So the series converges for \( \frac{{\left| x \right|}}{3} < 1,\)     A1

the radius of convergence is 3     A1

Note: Do not penalise lack of modulus signs.

[6 marks]

\({u_n} = \sqrt[3]{{{n^3} + 1}} - n\)

\( = n\left( {\sqrt[3]{{1 + \frac{1}{{{n^3}}} - 1}}} \right)\)     M1A1

\( = n\left( {1 + \frac{1}{{3{n^3}}} - \frac{1}{{9{n^6}}} + \frac{5}{{8{\text{l}}{n^9}}} - ... - 1} \right)\)     A1

using \({v_n} = \frac{1}{{{n^2}}}\) as the auxilliary series,     M1

since \(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{{v_n}}} = \frac{1}{3}{\text{ and }}\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + …\) converges     M1A1

then \(\sum {{u_n}} \) converges     A1

Note: Award M1A1A1M0M0A0A0 to candidates attempting to use the integral test.

[7 marks]

Some corners were cut in applying the ratio test and some candidates tried to use the comparison test. With careful algebra finding the radius of convergence was not too difficult. Often the interval of convergence was given instead of the radius.

Part (b) was done only by the best candidates. A little algebraic manipulation together with an auxiliary series soon gave the answer.

Some corners were cut in applying the ratio test and some candidates tried to use the comparison test. With careful algebra finding the radius of convergence was not too difficult. Often the interval of convergence was given instead of the radius.

Part (b) was done only by the best candidates. A little algebraic manipulation together with an auxiliary series soon gave the answer.