• October 26th 2009, 05:11 AM
Biscaim
Hello,

Show that $lim_{n\rightarrow +\infty}n[1-\frac{(n+1)^{n}}{en^{n}}]=\frac{1}{2}.$

• October 27th 2009, 06:19 PM
mr fantastic
Quote:

Originally Posted by Biscaim
Hello,

Show that $lim_{n\rightarrow +\infty}n[1-\frac{(n+1)^{n}}{en^{n}}]=\frac{1}{2}.$

It looks to me like you'll need to develop this expression as a series. See - Wolfram|Alpha
• October 27th 2009, 08:15 PM
rn443
Quote:

Originally Posted by Biscaim
Hello,

Show that $lim_{n\rightarrow +\infty}n[1-\frac{(n+1)^{n}}{en^{n}}]=\frac{1}{2}.$

$n[1-\frac{(n+1)^n}{en^n}] = \frac{1 - \frac{1}{e}\Big(1+\frac{1}{n}\Big)^n}{\frac{1}{n}}$. 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).