Show that $\displaystyle lim_{n\rightarrow +\infty}n[1-\frac{(n+1)^{n}}{en^{n}}]=\frac{1}{2}.$
Thanks in advance.
$\displaystyle n[1-\frac{(n+1)^n}{en^n}] = \frac{1 - \frac{1}{e}\Big(1+\frac{1}{n}\Big)^n}{\frac{1}{n}}$. By definition of e, the numerator approaches 1 - 1/e * e = 0, so maybe applying L'Hopital's rule will be successful (I haven't tried it).