Question

$\color{red} \displaystyle \rm \lim_{x \to \infty } \log \bigg( \sum \limits _{n = 0}^{x} \int_{0}^{x} \frac{ {y}^{n - 1} {e}^{ - \frac{y}{e} } }{e((n - 1)! {)}^{2} } dy\bigg )$​

Answer 1

$\color{red} \displaystyle \rm \lim_{x \to \infty } \log \bigg[ \sum \limits _{n = 0}^{x} \int_{0}^{x} \frac{ {y}^{n - 1} {e}^{ - \frac{y}{e} } }{e((n - 1)! {)}^{2} } dy\bigg]$

$\color{red} \displaystyle \rm \lim_{x \to \infty } \log \bigg[ \sum \limits _{n = 0}^{x} \int_{0}^{ \frac{x}{e} } \frac{ {e}^{n - 1} {u}^{n - 1} {e}^{ - u} }{ e((n - 1)! {)}^{2} } \: e \: du\bigg]$

$\color{red} \displaystyle \rm \lim_{x \to \infty } \log \bigg[ \sum \limits _{n = 0}^{x} \frac{ {e}^{n - 1} }{((n - 1)! {)}^{2} } \int_{0}^{ \frac{x}{e}} {e}^{ - u} \: {u}^{n - 1} \: du\bigg]$

$\color{red} \displaystyle \rm \lim_{x \to \infty } \log \bigg[ \sum \limits _{n = 0}^{x} \frac{ {e}^{n - 1} }{((n - 1)! {)}^{2} } \: (n - 1)!\bigg]$

$\color{red} \displaystyle \rm \lim_{x \to \infty } \log \bigg[ \sum \limits _{n = 0}^{x} \frac{ {e}^{n - 1} }{(n - 1)! {}^{} } \bigg]$

$\color{red} \displaystyle \rm \lim_{x \to \infty } \log \bigg[ \sum \limits _{n = 1}^{x} \frac{ {e}^{n } }{n! {}^{} } \bigg]$

$\color{red} \displaystyle \rm \lim_{x \to \infty } log( {e}^{e} ) = e$