For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n }$ is defined as in Q19. We assume $k \in I _ { n } = \left\{ k \in \llbracket 0 , n \rrbracket \mid x _ { n , k } \in [ 0 , \ell + 1 ] \right\}$. Deduce that, as $n$ tends to $+ \infty$, we have $$B _ { n } \left( x _ { n , k } \right) = \frac { 1 } { \sqrt { 2 \pi } } \frac { 1 + O \left( \frac { 1 } { n } \right) } { \left( \frac { 2 k } { n } \right) ^ { k + 1 / 2 } \left( 2 - \frac { 2 k } { n } \right) ^ { n - k + 1 / 2 } }$$
For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n }$ is defined as in Q19. We assume $k \in I _ { n } = \left\{ k \in \llbracket 0 , n \rrbracket \mid x _ { n , k } \in [ 0 , \ell + 1 ] \right\}$.
Deduce that, as $n$ tends to $+ \infty$, we have
$$B _ { n } \left( x _ { n , k } \right) = \frac { 1 } { \sqrt { 2 \pi } } \frac { 1 + O \left( \frac { 1 } { n } \right) } { \left( \frac { 2 k } { n } \right) ^ { k + 1 / 2 } \left( 2 - \frac { 2 k } { n } \right) ^ { n - k + 1 / 2 } }$$