[tex]S_n=\displaystyle\sum_{k=1}^n\frac1k[/tex]
You have
[tex]\ln e^{S_n}=S_n\ln e=S_n[/tex]
so showing that [tex]e^{S_n}>n+1[/tex] amounts to the same as showing that [tex]S_n>\ln(n+1)[/tex].
As [tex]n\to\infty[/tex], you have [tex]\ln(n+1)\to\infty[/tex]. By comparison, then, it follows that [tex]S_n\to\infty[/tex] at a faster rate, which means [tex]S_\infty[/tex] must diverge.
