TITLE: What is known about the 'Double log Eulers constant', $\lim_{n \to \infty}{\sum_{k=2}^n\frac{1}{k\ln{k}}-\ln\ln{n}}$?
QUESTION [6 upvotes]: 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 $\gamma$?

REPLY [2 votes]: Applying the Euler-Maclaurin Sum Formula we get
$$
\begin{align}
&\sum_{k=2}^n\frac1{k\log(k)}\\
&=\log(\log(n))+q+\frac1{2n\log(n)}-\frac1{12n^2}\left(\frac1{\log(n)}+\frac1{\log(n)^2}\right)\\
&+\frac1{720n^4}\left(\frac6{\log(n)}+\frac1{\log(n)^2}+\frac{12}{\log(n)^3}+\frac6{\log(n)^4}\right)\\
&-\frac1{15120n^6}\scriptsize\left(\frac{60}{\log(n)}+\frac{137}{\log(n)^2}+\frac{225}{\log(n)^3}+\frac{255}{\log(n)^4}+\frac{180}{\log(n)^5}+\frac{60}{\log(n)^6}\right)\\
&+\frac1{604800n^8}\left(\tiny\frac{2520}{\log(n)}+\frac{6534}{\log(n)^2}+\frac{13132}{\log(n)^3}+\frac{20307}{\log(n)^4}+\frac{23520}{\log(n)^5}+\frac{19320}{\log(n)^6}+\frac{10080}{\log(n)^7}+\frac{2520}{\log(n)^8}\right)\\
&-\frac1{1995840n^{10}}\left(\tiny\frac{15120}{\log(n)}+\frac{42774}{\log(n)^2}+\frac{97725}{\log(n)^3}+\frac{180920}{\log(n)^4}+\frac{269325}{\log(n)^5}+\frac{316365}{\log(n)^6}+\frac{283500}{\log(n)^7}+\frac{182700}{\log(n)^8}+\frac{75600}{\log(n)^9}+\frac{15120}{\log(n)^{10}}\right)\\
&+O\!\left(\frac1{n^{12}\log(n)}\right)
\end{align}
$$
If we use $n=10000$, we get $q$ to over $49$ places:
$$
\scriptsize\lim_{n\to\infty}\left(\sum_{k=2}^n\frac1{k\log(k)}-\log(\log(n))\right)=0.7946786454528994022038979620651495140649995908828
$$