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]