Solution
= Solution
For $m>n$, the $L^1$ difference is $\int_n^m x^{-1}dx=\log(m/n)$. Choosing $m=2n$ makes it $\log2$, so \b[the sequence is not Cauchy in $L^1$]. Its squared $L^2$ difference is $\int_n^m x^{-2}dx\leq1/n$, which tends uniformly to zero for $m>n$. Hence \b[it is Cauchy in $L^2$], with <norm> limit $x^{-1}$ on $(1,\infty)$.