Help with the limit $\lim_{n\to\infty}\frac{1}{\ln(\ln n)}\sum_{k=2}^n\frac{1}{k\ln k} $, most probably with Stolz-Cesaro theorem $$\lim_{n\to \infty}\frac{\frac{1}{2\ln2}+\frac{1}{3\ln3}+\ldots+\frac{1}{n\,\ln\,n}}{\ln(\ln\,n)}$$
The result is $1$ (according to the book, though it does not show the steps, which I'm interested in).
I've applied the theorem and it led me to an equally unhelpful limit.
 A: If you apply Stolz you get
$$\frac{1}{n\ln n(\ln \ln n-\ln\ln(n-1))}$$ Now take $e$ to the power of the denominator you get
$$\left(\frac{\ln n}{\ln(n-1)}\right)^{n\ln n}$$
Write this as 
$$\left(1+\frac{a_n}{n\ln n}\right)^{n\ln n}$$
Where $$a_n=n\ln n \frac{\ln (1+\frac{1}{n-1})}{\ln (n-1)}$$ its easy to see that $a_n\to 1$. 
And thus $$\left(1+\frac{a_n}{n\ln n}\right)^{n\ln n}\to e$$
Thus the denominator limits to $\ln e=1$ so the original expression above limits to $1$.
A: We apply the Stolz-Cesaro theorem: $$lim_{n\to \infty}\frac{(\frac{1}{2ln2}+\frac{1}{3ln3}+...+\frac{1}{n\,ln\,n}+\frac{1}{(n+1)\,ln(n+1)})-(\frac{1}{2ln2}+\frac{1}{3ln3}+...+\frac{1}{n\,ln\,n})}{ln(ln(n+1))-ln(ln\,n)}=lim_{n\to \infty}\frac{\,\frac{1}{(n+1)\,ln(n+1)}\,}{\,ln(\frac{ln(n+1)}{ln\,n})\,}$$
We know that:
$$ln\left(\frac{ln(n+1)}{ln\,n}\right)=ln\left(\frac{ln(n(1+\frac{1}{n}))}{ln\,n}\right)=ln\left(\frac{ln\,n+ln(1+\frac{1}{n})}{ln\,n}\right)=ln\left(1+\frac{ln(1+\frac{1}{n})}{ln\,n}\right)$$
Now,  $log(1+x)\ \sim x$, so $ln\left(1+\frac{ln(1+\frac{1}{n})}{ln\,n}\right) \sim \frac{ln(1+\frac{1}{n})}{ln\,n} \sim \frac{\frac{1}{n}}{ln\,n}=\frac{1}{n\,ln\,n}$ .
( "$\sim$" means "approximately equal to" in this context)
So: $$lim_{n\to \infty}\frac{\,\frac{1}{(n+1)\,ln(n+1)}\,}{\,ln(\frac{ln(n+1)}{ln\,n})\,}=lim_{n\to \infty}\frac{\,\frac{1}{(n+1)\,ln(n+1)}\,}{\,\frac{1}{n\,ln\,n}\,}=lim_{n\to \infty}\frac{n\,ln\,n}{(n+1)\,ln(n+1)}$$
which, in turn, is equal to $$\lim_{n\to \infty} \frac{n}{n+1}\cdot\lim_{n\to \infty}\frac{ln\,n}{ln(n+1)}=1\cdot1=1$$
So 1 is our final answer.
