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.

State the converse proposition.

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

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]

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

[1 mark]

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]