Proving $(1 + 1/n)^{n+1} \gt e$ I'm trying to prove that
$$ \left(1 + \frac{1}{n}\right)^{n+1} > e $$
It seems that the definition of $e$ is going to be important here but I can't work out what to do with the limit in the resulting inequality.
 A: Take logarithms of both sides: $(n+1) \log(1+1/n) > 1$.  With $t = 1/n$, this becomes
$\log(1+t) > 1/(1+1/t) = t/(1+t)$.  This is an equality at $t=0$, and the derivative of
$\log(1+t) - t/(1+t)$ is $t/(1+t)^2$, which is positive for $t \ge 0$. 
A: I would like to post a short proof of a stronger claim, namely that:

$$ \forall n\geq 1,\qquad \left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}\geq e.\tag{1} $$

By switching to logarithms, $(1)$ is equivalent to:
$$ \left(n+\frac{1}{2}\right)\int_{-1/2}^{1/2}\frac{dx}{n+\frac{1}{2}+x}\,dx \geq 1 \tag{2}$$
but $g(x)=\frac{1}{x+1}$ is a convex function on $\mathbb{R}^+$, hence $(2)$ trivially follows from the Hermite-Hadamard inequality.
A: $$
e^{-1} = \left(e^{-\frac{1}{n+1}}\right)^{n+1} > \left(1-\frac{1}{n+1}\right)^{n+1}
$$
and
$$
\left(1-\frac{1}{n+1}\right)^{n+1} \cdot \left(1 + \frac{1}{n}\right)^{n+1} = 1
$$
A: To complete The Chaz' answer:
You just need to show that the sequence $\bigl\{(1+{1\over n})^{n+1}\bigr\}$ is decreasing 
(one then easily shows its limit is $e$ if you know that $\bigl(1+{1\over n}\bigr)^n$ converges to $e$).
We use Bernoulli's inequality:
$$
(1+x)^n>1+nx,\quad \text{for }\ \  x>-1, n\ge 1.
$$
We have
$$
\eqalign{

{\bigl(1+{1\over n}\bigr)^{n+1}\over \bigl(1+ {1\over n+1}\bigr)^{n+1}}
&= \Bigl(1+{1\over n^2+2n}\Bigr)^{n+1}\cr
& >1+{n+1\over n^2+2n}\cr
& >1+{ 1\over n+1}\cr
&={n+2\over n+1}.
}
$$
Thus
$$
\Bigl(1+{1\over n+1}\Bigr)^{n+2} ={n+2\over n+1}\Bigl(1+{1\over n+1}\Bigr)^{n+1} < \Bigl(1+{1\over n}\Bigr)^{n+1}.
$$
And so the sequence $\bigl\{(1+{1\over n})^{n+1}\bigr\}$ is decreasing.
A: Can you show that $a_n = \left ( 1 + \frac{1}{n} \right ) ^{n + 1}$ (for $n = 1, 2, 3, ...$) is a decreasing sequence that converges to $e$  ?
Then 
$$\left ( 1 + \frac{1}{n} \right )^{n + 1} = \left ( 1 + \frac{1}{n} \right )^{1} \cdot\left ( 1 + \frac{1}{n} \right )^{n} $$
and taking limits (as $n \to \infty$) on both sides gives...
A: The following are lesser known facts, neverthless they are of some interest.
Let us introduce a tuning parameter $\alpha \in [0,\infty[$ and consider the sequence:
$$x_\alpha (n):=\left( 1+\frac{1}{n}\right)^{n+\alpha}\; .$$
Then $\displaystyle \lim_{n\to \infty} x_\alpha (n)= e$ for any $\alpha$, but the monotonicity and the position of $x_\alpha (n)$ with respect to $e$ changes with $\alpha$ (i.e., they both can be tuned by varying $\alpha$).
Then the following statements can be proved:

  
*
  
*If $1/2\leq \alpha $ then $x_\alpha (n)$ decreases strictly and converges to $e$ from above;
  
*There exists a number $a\in ]0,1/2[$ s.t.:
  
  
*
  
*if $0\leq \alpha < a$, then $x_\alpha (n)$ increases strictly and converges to $e$ from below; 
  
*if $a\leq \alpha <1/2$, then there exists $\nu =\nu(\alpha) \in \mathbb{N}$ s.t. $x_\alpha (n)$ decreases for $1\leq n\leq \nu$, increases for $n>\nu$ and converges to $e$ from below.
  
  

The number $a$ is something like $\ln 4 -1\approx 0.3863$. The proofs of these facts are tedious and lengthy but also elementary, for they rely on Differential Calculus.
Moreover, a simple computation with Taylor series expansion yields that $x_{1/2} (n)$ is the sequence which has the best rate of convergence to $e$ among the $x_\alpha$. In fact, we have:
$$\begin{split}
x_\alpha (n) -e&= \exp \left( (n+\alpha)\ \ln (1+1/n)\right) -e\\
&= \exp \left((n+\alpha) \left( \frac{1}{n}-\frac{1}{2n^2}+\text{o}(1/n^2)\right) \right) -e\\
&= \exp \left(1 +\frac{2\alpha -1}{2n} +\text{o}(1/n)\right) -e\\
&\approx \frac{e(2\alpha -1)}{2n} 
\end{split}$$
for $\alpha \neq 1/2$, but:
$$\begin{split}
x_{1/2} (n) -e&= \exp \left( (n+1/2)\ \ln (1+1/n)\right) -e\\
&= \exp \left((n+1/2) \left( \frac{1}{n}-\frac{1}{2n^2}+ \frac{1}{3n^3}+\text{o}(1/n^3)\right) \right) -e\\
&= \exp \left(1 +\frac{1}{12n^2} +\text{o}(1/n^2)\right) -e\\
&\approx \frac{e}{12n^2} \; .
\end{split}$$
