What is known about the 'Double log Eulers constant', $\lim_{n \to \infty}{\sum_{k=2}^n\frac{1}{k\ln{k}}-\ln\ln{n}}$?

Solution 1:

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 $$\int_{2}^\infty \frac{x^{-\delta-1}}{\log x}dx=-\log \delta -\gamma-\log \log 2+O\left(\delta\log(1/\delta)\right).$$

Solution 2:

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: $$ \scriptsize\lim_{n\to\infty}\left(\sum_{k=2}^n\frac1{k\log(k)}-\log(\log(n))\right)=0.7946786454528994022038979620651495140649995908828 $$