Proving of Boundedness and Monotonicity of a sequence $e(n) = (1 + \frac 1n)^n$ is convergent. How do we prove it?(proofs other than binomial expansion are welcome)
 A: We know that : 
$$\left(1+\frac{1}{n}\right)^n=\exp\left( n\ln\left( 1+\frac{1}{n} \right)\right).$$
Furthermore :
$$ \ln(1+x)=x+\underset{x\to 0}{o}(x).$$
Then you use that $\displaystyle \lim_{n\to +\infty} \frac{1}{n}=0$ to conclude.
A: By Bernoulli's inequality, we have $$\left(1-\frac 1{(n+1)^2}\right)^{n+1}\ge 1-\frac 1{n+1}=\frac n{n+1}$$ Now note that $$1-\frac 1{(n+1)^2}={n^2+2n\over n^2+2n+1}={n(n+2)\over (n+1)^2}=\frac {n+2}{n+1}\frac n{n+1},$$ so the above inequality reads $$\left(\frac {n+2}{n+1}\right)^{n+1}\left(\frac n{n+1}\right)^{n+1}\ge \frac n{n+1}$$ Multiplying both sides with $\displaystyle\left(\frac {n+1}n\right)^{n+1}$ gives you the monotonicity of the sequence.
To prove boundedness, use the binomial theorem to show $\left(1+\frac 1n\right)^n\le 3$ noting that $${n\choose k}\left(\frac 1n\right)^k=\frac 1{k!}{n(n-1)...(n-k+1)\over n^k}=\frac 1{k!} \frac{n}n\frac{n-1}n...\frac{n-k+1}n\le \frac 1{k!}\le \left(\frac 12\right)^{k-1}.$$
A: Consider $n≥2$ , then $1- \frac1{n}>0$ , by A.M.-G.M. inequality we get , 
$\frac { \{(1- \frac1{n})+(1- \frac1{n})+...n times\} +1 }{n+1}>\{1×(1- \frac1{n})^n\}^{1/{n+1}}$ , strict inequality holds as $1- \frac1{n}≠1$ , hence 
$ \frac {n-1+1}{n+1}>\{(1- \frac1{n})^n\}^{1/{n+1}}$ , that is $(1- \frac1{n+1})^{n+1}>(1- \frac1{n})^n$ holds for $n≥2$ ; whence the 
sequence $s_n=(1- \frac1{n})^n$ is strictly increasing for $n≥2$ ; hence $s_n=(1- \frac1{n})^n > s_2$ , for $n>2$ 
that is for $n>2$ , $(1- \frac1{n})^n > (1-\frac1{2})^2= \frac1{4}$. Now we know $\space$ $1> 1-\frac1{n^2}=(1+ \frac1{n})(1- \frac1{n})$ ; as
for $n>2$ , both $1+ \frac1{n} , 1- \frac1{n}$ are positive , by dividing both sides by $(1+ \frac1{n})$ and raising both 
sides to the power $n$ we get , $\frac 1{(1+ \frac1{n})^n}>(1- \frac1{n})^n > \frac1{4}$ i.e. $(1+ \frac1{n})^n < 4$ , for $n>2$ and it can be 
easily checked that for $n=1,2 $ ; $(1+ \frac1{n})^n < 4$ , hence $e(n) = (1 + \frac 1n)^n$ is bounded above. It can
be easily shown that $e(n)$ is increasing indeed by A.M.-G.M. inequality  , 
$\frac { \{(1+ \frac1{n})+(1+ \frac1{n})+...n times\} +1 }{n+1}>\{1×(1+ \frac1{n})^n\}^{1/{n+1}}$ , strict inequality holds as $1+ \frac1{n}≠1$ that is
$ \frac {n+1+1}{n+1}>\{(1+ \frac1{n})^n\}^{1/{n+1}}$ , hence $(1+ \frac1{n+1})^{n+1}>(1+ \frac1{n})^n$ holds for all positive integer $n$ , so 
$e(n)$ is increasing whence it is bounded below by $e(1)$ , moreover it is bounded above as we showed each term is less than $4$ , hence by monotone convergence theorem $e(n)$ is convergent. 
