Prove $(1 + \frac{1}{n})^n$ is bounded above I've checked similar questions on the site but couldn't find satisfactory solutions or hints. 
Also, is there a more general approach to proving whether a given sequence is bounded below or above?
 A: By the binomial formula:
\begin{eqnarray*}
\left(1+\frac{1}{n}\right)^n=1+1+\sum_{k=2}^n\binom{n}{k}\cdot\left(\frac{1}{n}\right)^k.
\end{eqnarray*}
Notice that
\begin{eqnarray*}
\binom{n}{k}\cdot\left(\frac{1}{n}\right)^k=\frac{n(n-1)\cdots(n-k+1)}{n^k}\cdot\frac{1}{k!}<\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}.
\end{eqnarray*}
So we get
\begin{eqnarray*}
\left(1+\frac{1}{n}\right)^n<2+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=3-\frac{1}{n}<3.
\end{eqnarray*}
A: Hint: Use the binomial formula:
$$\left(1 + \frac{1}{n}\right)^n =\sum_{j=0}^n\frac1{j!}\frac{n!}{(n-j)!}\frac1{n^j}= 1+1+\sum_{j=2}^n\frac1{j!} \prod_{k=1}^{j-1}(1-k/n) < \ldots$$
A: My question here
(What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists?)
has an elementary proof
that,
if $a_n = (1+1/n)^n$
and
$b_n = (1+1/n)^{n+1}$
then the
$a_n$ are increasing
and the
$b_n$ are decreasing.
Since
$a_n < b_n$,
all the $a_n$
are less than
any of the $b_n$.
Since
$b_5 = 46656/15625 < 3$,
all the $a_n < 3$.
A: Hint:  It is well-known that the limit is $e $.  It is also well-known that a convergent sequence is bounded...
I've posted this on this site before,  but couldn't find it.  So here goes again:  using a single interval upper and lower Riemann sum for $\ln x:=\int_1^x\dfrac 1t\operatorname dt$,  we get $\dfrac x{n+x}\le\ln(1+\frac xn)\le\dfrac xn$.
So, $e^{\frac x{n+x}}\le 1+\dfrac xn\le e^{\frac xn}$.
Thus $e^{\frac {nx}{n+x}}\le (1+\dfrac xn)^n\le e^x$.
Now let $n\to \infty$, for the proof of convergence.
I believe I first saw this proof in Best and Penner's Calculus.
A: Just for the heck of it.
If $x > 0$
then
$\begin{array}\\
\ln(1+x)
&=\int_0^x \dfrac{dt}{1+t}\\
&\lt x\\
\text{so}\\
n\ln(1+\frac1{n})
&\lt n(\frac1{n})\\
&=1\\
\text{so}\\
(1+\frac1{n})^n
&\lt e\\
\end{array}
$
