# 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 $\gamma$?

• you could calculate it to a certain precision and ask the inverse symbolic calculator... – draks ... Dec 21 '15 at 12:04
• @draks...: This recommendation is given a lot, but has anyone ever had much success with the inverse symbolic calculator? I have used that thing close to 100 times, and have only had it work once. – Eric Naslund Dec 21 '15 at 13:41
• @EricNaslund: I guess using the ISC is a matter of luck and the context of your numerical value. It's given me many results already, a few of which are values that turn out to be $\ln 2$ (ex 1), or $\ln(\phi)$ (ex 2), or $\phi^\sqrt{5}$ (ex 3). – Tito Piezas III Dec 21 '15 at 15:20
• @Tito Piezas: It might just be that I am frequently searching for constants like the one the OP is asking about that do not have a simple form. Finding $\phi^{\sqrt{5}}$ is pretty impressive – Eric Naslund Dec 21 '15 at 15:50
• @EricNaslund: Yes, that one was pretty fortunate. :) – Tito Piezas III Dec 21 '15 at 15:53

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

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

• I'm not entirely sure, but I think it would be more efficient to apply the Euler-Maclaurin summation formula to $\sum_{k=a}^n\frac1{k\ln(k)}$ for larger choices of $a$ and manually adding the sum over $k\in[2,a)$. – Simply Beautiful Art Sep 25 '17 at 18:14
• @SimplyBeautifulArt: not quite sure what you are saying. to get $49$ places of $q$, I summed $\frac1{k\log(k)}$ (manually?) to $k=10000$ and subtracted the asymptotic expansion with $n=10000$. – robjohn Sep 25 '17 at 18:22
• I meant this: $$\sum_{k=1}^nf(k)=\sum_{k=1}^{a-1}f(k)+\sum_{k=a}^nf(k)=\sum_{k=1}^{a-1}f(k)+\int_a^nf(x)~\mathrm dx+\frac{f(a)+f(n)}2+\dots+R_p$$I've found that applying the EMF this way lowers the remainder term by a lot. Of course, use the related function here. – Simply Beautiful Art Sep 25 '17 at 18:26
• See this answer for another example of what I mean. In this link, note that choosing small values of $n$ will slow the rate of convergence. – Simply Beautiful Art Sep 25 '17 at 18:32
• I don't see how this is different from what I am doing. One has to balance the number of asymptotic terms with the number of manual terms. Too many asymptotic terms and not enough manual terms and the asymptotic terms cause inaccuracy. If $n$ is not large enough, the asymptotic terms can start increasing if we take too many. – robjohn Sep 25 '17 at 18:43