limit and sequence

**bearej50** · April 5th 2009, 11:54 AM

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)

**Chris L T521** · April 5th 2009, 11:59 AM

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)

$\begin{aligned}\lim \left(\frac{n+2}{n+1}\right)^{n+3}&= \lim\left(\frac{n+1+1}{n+1}\right)^{n+3}\\ &= \lim\left(1+\frac{1}{n+1}\right)^{n+3}\\&=\lim\lef t(1+\frac{1}{n+1}\right)^{\left(n+1\right)\left(\f rac{n+3}{n+1}\right)}\\ &= e^{\lim\frac{n+3}{n+1}}\\&=\boxed{e}\end{aligned}$

**Jhevon** · April 5th 2009, 12:00 PM

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 $s_n = \left( 1 + \frac 1{n + 1} \right)^{(n + 1) + 2} = \left( 1 + \frac 1{n + 1} \right)^2 \cdot \left( 1 + \frac 1{{\color{red}n + 1}} \right)^{\color{red}n + 1}$

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

Thread: limit and sequence

LinkBack

Thread Tools

Display

limit and sequence

Similar Math Help Forum Discussions

limit of sequence

limit of function of a sequence equals the function of the limit of that sequence?

limit of a sequence

Limit of a Sequence

Limit of sequence

Search Tags