Question

$\large\displaystyle \red{ { \rm{{\lim_{n \to \infty } {e}^{ - n} \sum_{ \mathbb{k = 0}}^{n} \frac{ {n}^{ \mathbb{k}} }{ \mathbb{k!}} }}}}$​

Answer 1

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left [{e}^{ - n} \sum_{ {k = 0}}^{n} \frac{ {n}^{ {k}} }{ {k!}} \right] }}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \bigg[ {e}^{ - n} \sum_{ {k = 0}}^{n} exp(k \ln(n) - \ln(k!)) \bigg]}}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left \{{e}^{ - n} \sum_{ {k = 0}}^{n} exp \bigg(n \ln(n) - \ln(n!) - \frac{ {1}^{ {}} }{ {2n}} [k - n]^{2} \bigg)\right \} }}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left \{ \frac{ {e}^{ - n} {n}^{n} }{n!} \int_{0}^{ n} exp \bigg( - \frac{1}{2n} [k - n] {}^{2} \bigg)dk \right \} }}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left [ \frac{ {e}^{ - n} {n}^{n} }{n!} \int_{ - n}^{0} exp \bigg( - \frac{ {k}^{2} }{2n} \bigg)dk\right] }}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left [ \frac{ {e}^{ - n} {n}^{n} }{n!}\sqrt{2n} \int_{ - \frac{ \sqrt{n} }{2} }^{0} exp( - {k}^{2} )dk\right] }}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left [ \frac{ \sqrt{2} {n}^{n + \frac{1}{2} } {e}^{ - n} }{n!} \int_{ - \infty }^{0} exp( - {k}^{2} )dk\right] }}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left [ \frac{ \sqrt{2} {n}^{n + \frac{1}{2} } {e}^{ - n} }{n!} \frac{ \sqrt{\pi} }{2} \right] }}}}$

$\displaystyle \red{ { \sf{ \frac{1}{2} {\lim_{n \to \infty } \left [ \frac{ \sqrt{2\pi} {n}^{n + \frac{1}{2} } {e}^{ - n} }{n!} \right] }}}}$

$\huge\red{ \frac{1}{2} }$

Answer 2

Answer:

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left [{e}^{ - n} \sum_{ {k = 0}}^{n} \frac{ {n}^{ {k}} }{ {k!}} \right] }}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \bigg[ {e}^{ - n} \sum_{ {k = 0}}^{n} exp(k \ln(n) - \ln(k!)) \bigg]}}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left \{{e}^{ - n} \sum_{ {k = 0}}^{n} exp \bigg(n \ln(n) - \ln(n!) - \frac{ {1}^{ {}} }{ {2n}} [k - n]^{2} \bigg)\right \} }}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left \{ \frac{ {e}^{ - n} {n}^{n} }{n!} \int_{0}^{ n} exp \bigg( - \frac{1}{2n} [k - n] {}^{2} \bigg)dk \right \} }}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left [ \frac{ {e}^{ - n} {n}^{n} }{n!} \int_{ - n}^{0} exp \bigg( - \frac{ {k}^{2} }{2n} \bigg)dk\right] }}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left [ \frac{ {e}^{ - n} {n}^{n} }{n!}\sqrt{2n} \int_{ - \frac{ \sqrt{n} }{2} }^{0} exp( - {k}^{2} )dk\right] }}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left [ \frac{ \sqrt{2} {n}^{n + \frac{1}{2} } {e}^{ - n} }{n!} \int_{ - \infty }^{0} exp( - {k}^{2} )dk\right] }}}}$

$\displaystyle \red{ { \sf{{\lim_{n \to \infty } \left [ \frac{ \sqrt{2} {n}^{n + \frac{1}{2} } {e}^{ - n} }{n!} \frac{ \sqrt{\pi} }{2} \right] }}}}$

$\displaystyle \red{ { \sf{ \frac{1}{2} {\lim_{n \to \infty } \left [ \frac{ \sqrt{2\pi} {n}^{n + \frac{1}{2} } {e}^{ - n} }{n!} \right] }}}}$

$\huge\red{ \frac{1}{2} }$