Show that, for $n \ge 2$ , ${S_{2n}} > {S_n} + \frac{1}{2}$ .

Deduce that ${S_{2m + 1}} > {S_2} + \frac{m}{2}$ .

Hence show that the sequence $\left\{ {{S_n}} \right\}$ is divergent.

${S_{2n}} = {S_n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} +  \ldots  + \frac{1}{{2n}}$     M1

$> {S_n} + \frac{1}{{2n}} + \frac{1}{{2n}} +  \ldots  + \frac{1}{{2n}}$     M1A1

$= {S_n} + \frac{1}{2}$     AG

[3 marks]

Replacing $n$ by $2n$,

${S_{4n}} > {S_{2n}} + \frac{1}{2}$     M1A1

$> {S_n} + 1$     A1

Continuing this process,

${S_{8n}} > {S_n} + \frac{3}{2}$     (A1)

In general,

${S_{{2^m}n}} > {S_n} + \frac{m}{2}$     M1A1

Putting $n = 2$     M1

${S_{{2^{m + 1}}}} > {S_2} + \frac{m}{2}$     AG

[7 marks]

Consider the (large) number $N$.     M1

Then, ${S_{2m + 1}} > N$ if ${S_2} + \frac{m}{2} > N$     A1

i.e. if $m > 2(N - {S_2})$     A1

This establishes the divergence.     AG

[3 marks]