limit and sequence

• Apr 5th 2009, 11:54 AM
bearej50
limit and sequence
The limit, as n approaches infinity, of (1 + (1/n))^n = e (by e, I mean 2.71828). Use this to find the limit of the following sequence:

sn = ((n+2)/(n+1))^(n+3)
• Apr 5th 2009, 11:59 AM
Chris L T521
Quote:

Originally Posted by bearej50
The limit, as n approaches infinity, of (1 + (1/n))^n = e (by e, I mean 2.71828). Use this to find the limit of the following sequence:

sn = ((n+2)/(n+1))^(n+3)

\displaystyle \begin{aligned}\lim \left(\frac{n+2}{n+1}\right)^{n+3}&= \lim\left(\frac{n+1+1}{n+1}\right)^{n+3}\\ &= \lim\left(1+\frac{1}{n+1}\right)^{n+3}\\&=\lim\lef t(1+\frac{1}{n+1}\right)^{\left(n+1\right)\left(\f rac{n+3}{n+1}\right)}\\ &= e^{\lim\frac{n+3}{n+1}}\\&=\boxed{e}\end{aligned}
• Apr 5th 2009, 12:00 PM
Jhevon
Quote:

Originally Posted by bearej50
The limit, as n approaches infinity, of (1 + (1/n))^n = e (by e, I mean 2.71828). Use this to find the limit of the following sequence:

sn = ((n+2)/(n+1))^(n+3)

note that what you have is $\displaystyle s_n = \left( 1 + \frac 1{n + 1} \right)^{(n + 1) + 2} = \left( 1 + \frac 1{n + 1} \right)^2 \cdot \left( 1 + \frac 1{{\color{red}n + 1}} \right)^{\color{red}n + 1}$

now, do you see how to use what you were given?