Using the recurrence relation $K _ { n } = K _ { n + 1 } + \frac { 1 } { 2 n } K _ { n }$ and the fact that $I_n \sim K_n$, deduce a simple equivalent of $I _ { n }$ as $n$ tends to $+ \infty$, where $I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$.