$\lim\limits_{n\to\infty}$ $(1+ {2\over n})^n$ = $e^2$ 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.

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.
