How to find the limit of $\dfrac{\ln(\ln(\frac{n}{n-1}))}{\ln(n)}$? How do you find 
$$\lim_{n \to\infty} \dfrac{\ln(\ln(\frac{n}{n-1}))}{\ln(n)}$$
I know it's $-1$, but I had to plot it.
 A: Hint: You can use L'hopital's rule twice.
Added: If you use L'hopital's rule twice then you get the expression
$$ -\frac{n}{n-1}. $$ 
Another approach:
$$  \dfrac{\ln(\ln(\frac{n}{n-1}))}{\ln(n)} =  \dfrac{\ln(-\ln(1-\frac{1}{n}))}{\ln(n)}= \dfrac{\ln(-(-\frac{1}{n}-O(\frac{1}{n^2})))}{\ln(n)}\sim \frac{\ln(\frac{1}{n})}{\ln(n)}=-\frac{\ln(n)}{\ln(n)}=-1.$$
Note that, in the above derivations, we used the Taylor series of $$\ln(1-x)=-x-\frac{x^2}{2}-\dots\,,$$
and the following property of $\ln(x)$
$$ \ln(1/a) = -\ln(a). $$ 
A: Hint:
What is $\lim_{n \to \infty}  \dfrac{n}{n-1}$ ?
Convert the expression into $\dfrac{\pm \infty}{\infty}$ form. Then you have L'hoptial's rule. 
A: If $0\le x\le 1$, then $x\ln 2 \le \ln(1+x)\le x$. (This inequality follows that $f(x)=\ln(1+x)$ is concave function.) Since $x\mapsto \ln(1+x)$ is order-preserving, so
$$\frac{\ln\ln\left(1+\frac{1}{n-1}\right)}{\ln n}\le \frac{\ln \frac{1}{n-1}}{\ln n}=-\frac{\ln (n-1)}{\ln n}$$
for all $n\in\mathbb{N}$ and
$$\frac{\ln\ln\left(1+\frac{1}{n-1}\right)}{\ln n}\ge \frac{\ln \left( \frac{1}{n-1}\ln 2\right)}{\ln n}=-\frac{\ln (n-1)-\ln 2}{\ln n}$$
for all $n\in\mathbb{N}$. Therefore
$$-\frac{\ln (n-1)-\ln 2}{\ln n}\le \dfrac{\ln(\ln(\frac{n}{n-1}))}{\ln(n)} \le- \frac{\ln (n-1)}{\ln n}$$
and take $n\to\infty$ each side of inequality then we get desired result.
A: Here's a different approach. Let $n=e^m$ and look at the limit a $m\to\infty$. You get
$$
\frac{\log\log\frac{e^m}{e^m-1}}{\log e^m}=\frac{\log\log\left(1+\frac{1}{e^m-1}\right)}{m}
$$
Since clearly $\frac{1}{e^m-1}\to0$ you can use the approximation $\log(1+x)\approx x$. Then you have
$$
\frac{\log\left(\frac{1}{e^m-1}\right)}{m}=\frac{-\log(e^m-1)}{m}\approx\frac{-m}{m}=-1
$$
