What is known about the 'Double log Eulers constant', $\lim_{n \to \infty}{\sum_{k=2}^n\frac{1}{k\ln{k}}-\ln\ln{n}}$? The Euler constant is defined as $$\gamma = \lim_{n \to \infty}{\sum_{k=1}^n\frac{1}{k}-\ln{n}}$$
Let $$q = \lim_{n \to \infty}{\sum_{k=2}^n\frac{1}{k\ln{k}}-\ln\ln{n}}$$
I managed to prove that $$\frac{1}{3\ln{3}}+\frac{1}{2\ln{2}}-\ln\ln{3} \geq q \geq \frac{1}{2\ln{2}}-\ln\ln{3}$$
Is there something known about the constant $q$? For instance, is $q$ expressible in terms of $\gamma$?
 A: In some sense this constant is the negative first Stieltjes constant, although just like the Stieltjes constant $\gamma_{1}$, not too much can be said about it. We have that 
$$\sum_{k=2}^{N}\frac{1}{k\log k}=\log\log N+K+O\left(\frac{1}{N}\right).
 $$ where $$K=\frac{1}{\log4}-\log\log2-\int_{2}^{\infty}\{x\}\frac{\log x+1}{x^{2}\log^{2}x}dx.$$ 
The Stieltjes constants, defined by $$\gamma_{n}=\lim_{N\rightarrow\infty}\sum_{k=1}^{N}\frac{\left(\log k\right)^{n}}{k}-\frac{\left(\log N\right)^{n+1}}{n+1}
 $$ cannot be written in terms of $\gamma_0$ or other known constants, and in some sense your constant $K$ is $\gamma_{-1}$, the $-1^{st}$ Stieltjes constant. 
The above equation for $K$ can be proven using partial summation: rewriting the integral as a Riemann Stieltjes integral we have that $$\sum_{k=2}^{N}\frac{1}{k\log k} = \int_{2}^N \frac{1}{x\log x} d\lfloor x\rfloor =\int_{2}^{N}\frac{1}{x\log x}dx-\int_{2}^{N}\frac{1}{x\log x}d\{x\}$$ $$
 = \int_{2}^{N}\frac{1}{x\log x}dx-\frac{\{x\}}{x\log x}\biggr|_{2-}^{N}-\int_{2}^{N}\{x\}\frac{\log x+1}{x^{2}\log^{2}x}dx$$ and this last line simplifies to become $$\log \log N+\frac{1}{\log4}-\log\log2-\int_{2}^{\infty}\{x\}\frac{\log x+1}{x^{2}\log^{2}x}dx+O\left(\frac{1}{n}\right).$$



Side note: This is not that related, but I wanted to note that by examining this sum, we can reprove some of what did wrote in this answer, and we can show that as $s\rightarrow 0$ $$\int_2^\infty \frac{x^{-s-1}}{\log x}dx=-\log(s)-\gamma-\log \log 2+O(s\log(s)).$$ I am including this in the answer because I think it puts things into a greater context. 
Let $\Lambda(n)$ be the Von Mangoldt Lambda function, and $\gamma_0$ the Euler-Mascheroni Constant. Then we have the expansion of the similar sum $$\sum_{n\leq x}\frac{\Lambda(n)}{n\log n}=\log\log x+\gamma_{0}+O\left(\frac{1}{\log x}\right),$$ which appears in the proof of theorem 2.7 in Montgomery and Vaughn. Let $$S(x)=\sum_{2\leq k\leq x}\frac{1}{k\log k}- \sum_{n\leq x}\frac{\Lambda(n)}{n\log n},$$ and examine $I=\delta \int_1^\infty S(x)x^{-\delta -1}dx$ as $\delta\rightarrow 0$. As $S(x)=(K-\gamma_0)+O(1/\log x)$, it follows that $I=K-\gamma_0+O(\delta \log (1/\delta)$. Then, since $$\sum_{n=1}^{\infty}a_{n}n^{-s}=s\int_{1}^{\infty}A(x)x^{-s-1}dx$$ (Theorem 1.3 of Montgomery and Vaughn) we see that $$\sum_{n=2}^{\infty}\frac{n^{-\delta}}{n\log n}-\log \zeta(\delta+1)=O(\delta\log(1/\delta)$$ as $\delta\rightarrow 0$, and so $$\sum_{n=2}^\infty \frac{n^{-\delta-1}}{\log n}=-\log \delta+(K-\gamma)+O(\delta\log(1/\delta).$$ Now, writing the left hand side as a Riemann Stieltjes integral allows us to conclude that $$\int_{2}^\infty \frac{x^{-\delta-1}}{\log x}dx=-\log \delta -\gamma-\log \log 2+O\left(\delta\log(1/\delta)\right).$$
