$\lim\limits_{n\to\infty}$ $(1+ {2\over n})^n$ = $e^2$ If I set N=$2\over n$, the equation becomes $(1+ N)^{2\over N}$, which I can take the natural log of and work down until I reach $e^2$. However, my teacher wants me to use subsequences, starting with $x_n$ = $(1+ {1\over n})^n = e$, to prove this and I am stuck.
 A: You could rewrite this as 
$$\lim_{n \to \infty}\left(1+\frac{1}{n/2}\right)^n = \lim_{n \to \infty}\left(\left(1+\frac{1}{n/2}\right)^{n/2}\right)^2 = \lim_{n\to\infty} (x_n)^2$$
where $x_n=\left(1+\frac{1}{n/2}\right)^{n/2}$. Now as $n \to \infty$, $\frac{n}{2} \to \infty$ as well. Hence $\lim_{n\to\infty} x_n = e$. Thus, bringing the limit inside the square (which we can do since squaring is continuous):
$$\lim_{n \to \infty}\left(1+\frac{1}{n/2}\right)^n = \lim_{n \to \infty}(x_n)^2= \left(\lim_{n\to\infty} x_n \right)^2=e^2.$$
This doesn't "use subsequences," but it makes use of the sequence whose limit you already know, if that was your teacher's intention.
A: Let y=(1+2/n)^n
ln⁡y=ln⁡〖(1+2/n)^n 〗
lim┬(n→∞)⁡ln⁡〖y= lim┬(n→∞)⁡ln⁡〖(1+2/n)^n 〗=ln⁡lim┬(n→∞)⁡y 
〖lim┬(n→∞)  n〗⁡ln⁡(1+2/n) =∞×0;hence undefined.
Proceed by using L’Hopitals Rule:
〖lim┬(n→∞)  n〗⁡ln⁡(1+2/n) =lim┬(n→∞)⁡〖ln⁡(1+2/n)/(1/n)〗
                         =lim┬(n→∞)⁡〖((-2)/(n^2 (1+2/n) ))/((-1)/n^2 )〗                             =lim┬(n→∞)⁡〖2/((1+2/n) )〗→2     
Hence ln⁡lim┬(n→∞)⁡y =2=ln⁡〖e^2 〗. Hence lim┬(n→∞)⁡〖(1+2/n)^n 〗=e^2
A: Let $a_n$  =$(1+2/n)^n$ and  $a_{n+1}$=$1+$($\frac{2}{n+1})^{n+1} $
Proceed by their binomial expansion.
$a_n$=$1+n$($\frac{2}{n})$+n($\frac{n-1}{2!})$ $(\frac{2}{n})^{2}$+⋯+$\frac{n(n-1)(n-2)…}{n!}$ ($\frac{2}{n})^{n}$
$a_n=$1+2+($\frac{1}{2!})$ $2^2$(1-$\frac{1}{n})$+($\frac{1}{3!})$ $2^3$(1-$\frac{1}{n})$(1-$\frac{2}{n})$+⋯+($\frac{1}{n!})$ $2^n$(1-$\frac{1}{n})$(1-$\frac{2}{n})$...(1-$\frac{n-1}{n})$
$a_{n+1}=$1+2+($\frac{1}{2!})$ $2^2$(1-$\frac{1}{n+1})$+($\frac{1}{3!})$ $2^3$(1-$\frac{1}{n+1})$(1-$\frac{2}{n+1})$+⋯+($\frac{1}{{n+1}!})$ $2^n$(1-$\frac{1}{n+1})$(1-$\frac{2}{n+1})$...(1-$\frac{n}{n+1})$
For n≥2,$a_{n+1}$ >a_n Hence Monotone increasing.
$a_n  ≤ 1+2+(2/2!)+(2/3!)+⋯
a_n  ≤1+(2+(2/2^2) +(2/2^3) +⋯)
a_n  ≤ 1+2(1+1/2^2 +1/2^3 +⋯)
By using sum to infinity, $a_n ≤1+2(2(1-1/2^n ))
$a_n ≤1+2(2-2^(1-n) )≤1+4-2^(2-n)≤5;as n→∞.
Hence convergent and bounded above by 5.
