Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 24J / i / Solution

For $m>n$, the $L^1$ difference is ${\int }_n^m{x}^{-1}dx=\log (m/n)$. Choosing $m=2n$ makes it $\log 2$, so the sequence is not Cauchy in $L^1$. Its squared $L^2$ difference is ${\int }_n^m{x}^{-2}dx\le 1/n$, which tends uniformly to zero for $m>n$. Hence it is Cauchy in $L^2$, with norm limit $x^{-1}$ on $(1,\infty )$.