Show that $a(n) = (1/n)^{1+1/n}$ is monotonically decreasing $a(n)$ tends to $0$, as $n$ tends to $\infty$, but I am having trouble showing $a(n) > a(n+1)$. 
I tried to use ln (n+1) - ln n >= 1/(n+1). so ln (n+1) >= ln n +1/(n+1) =>
1/(e^(ln n + 1/(n+1)) >= 1/(e^ln (n+1)) =>
1/(e^ln n)^(1+1/(n+1)) >= 1/(e^(ln n + 1/(n+1))^(1+1/(n+1)) >= 1/(e^ln (n+1))^(1+1/(n+1)). I think. 
I tried some additional ideas like multiplying both sides by 1/(e^ln n)^(1/(n^2 + n))
 but could not get the inequality I needed. 
 A: Define $$f(x)= -(1+   \frac{1}{x})\ln x .$$
Then $$f'(x)=  \frac{1}{x^2}\ln x -\frac{1}{x} -\frac{1}{x^2}= \frac{  \ln x -x-1}{x^2}$$
But using tha $e^x\geq x$ we get $x \geq \ln x \Rightarrow 0>-1\geq\ln x -x-1$.
Therefore $f(x)$ in decreasing, now notice that 
$$a_n=e^{f(\dfrac{1}{n})}$$ and so we conclude.
A: All that is needed is
$(1+1/n)^{n+1}
> e
$
and
$e^n > n$,
which is easily proved by induction assuming
$e > 2$.
If
$a(n) = (1/n)^{1+1/n}
$,
then
$\begin{array}\\
a(n) > a(n+1)\\
\iff\\
(1/n)^{1+1/n} > (1/(n+1))^{1+1/(n+1)}\\
\iff\\
(1/n)^{(n+1)/n} > (1/(n+1))^{(n+2)/(n+1)}\\
\iff\\
(1/n)^{(n+1)^2} > (1/(n+1))^{n(n+2)}\\
\iff\\
n^{(n+1)^2} < (n+1)^{n(n+2)}\\
\iff\\
n^{n^2+2n+1} < (n+1)^{n(n+2)}\\
\iff\\
n < (1+1/n)^{n(n+2)}\\
\end{array}
$
But
$\begin{array}\\
(1+1/n)^{n(n+2)}
&>(1+1/n)^{n(n+1)}\\
&>e^{n}
\qquad\text{since }(1+1/n)^{n+1}>e\\
&> n
\end{array}
$
Therefore
$a(n) > a(n+1)$.
A: Similar to clark's answer, consider $$f(x)=\left(\frac{1}{x}\right)^{\frac{1}{x}+1}$$ $$f'x)=-\left(\frac{1}{x}\right)^{\frac{1}{x}+3} \left(x-\log(x)
   +1\right)$$ The derivative cancels when $x=-W(-e)$ which is a complex number ($W(z)$ being Lambert function); so the derivative is always negative and $f(x)$ is a decreasing function. 
For large values of $x$, you could find that $$f(x)=\frac 1x -\frac{\log(x)}{x^2}+O\left(\frac{1}{x^3}\right)$$
