Finding the limit of $\left(\sum\limits_{k=2}^n\frac1{k\log k}\right)-\left(\log \log n\right)$ There is this well known limit:
$$\lim_n \sum_{k=1}^n \frac 1k -\log n=\gamma$$
Where $\log$ is the natural logarithm and $\gamma$ is Euler constant.
I was wondering if my similar situation yelds to a similar result:
$$\lim_n \sum_{k=2}^n \frac 1{k\log k}-\log \log n=?$$
I know there is a formula (always due to Euler) which can be used in those situations but I can't see if there is a way to put the result in "closed form" rather than having only an approximation (supposing the limit exists and is finite at first)
 A: Using $\log\left(1+\frac1x\right)\ge\frac1{x+1}$, we get
$$
\begin{align}
&\left(\sum_{k=2}^{n+1}\frac1{k\log(k)}-\log(\log(n+1))\right)-\left(\sum_{k=2}^n\frac1{k\log(k)}-\log(\log(n))\right)\\
&=\frac1{(n+1)\log(n+1)}-\log\left(\frac{\log(n+1)}{\log(n)}\right)\\
&\le\frac1{(n+1)\log(n+1)}-\log\left(\frac{\log(n)+\frac1{n+1}}{\log(n)}\right)\\
&=\frac1{(n+1)\log(n+1)}-\log\left(1+\frac1{(n+1)\log(n)}\right)\\
&\le\frac1{(n+1)\log(n+1)}-\frac1{(n+1)\log(n)+1}\\
&\le\frac1{(n+1)\left(\log(n)+\frac1{n+1}\right)}-\frac1{(n+1)\log(n)+1}\\
&=0
\end{align}
$$
Therefore,
$$
\sum_{k=2}^n\frac1{k\log(k)}-\log(\log(n))
$$
is decreasing. Furthermore,
$$
\begin{align}
\sum_{k=2}^n\frac1{k\log(k)}-\log(\log(n))
&\ge\int_2^{n+1}\frac{\mathrm{d}x}{x\log(x)}-\log(\log(n))\\
&=\log(\log(n+1))-\log(\log(2))-\log(\log(n))\\[9pt]
&\ge-\log(\log(2))
\end{align}
$$
is bounded below,
Thus,
$$
\lim_{n\to\infty}\left(\sum_{k=2}^n\frac1{k\log(k)}-\log(\log(n))\right)
$$
exists.
In fact, in this answer, this limit is computed to $49$ places using the Euler-Maclaurin Sum Formula:
$$
0.7946786454528994022038979620651495140649995908828
$$
A: Bit long for a comment, not really a finished answer.

More generally, consider
$$I(s)=\int_2^\infty\left(\frac1{\lfloor x\rfloor^s\ln\lfloor x\rfloor}-\frac1{x^s\ln(x)}\right)~\mathrm dx$$
By differentiating under the integral,
$$I'(s)=\int_2^\infty\left(\frac1{\lfloor x\rfloor^s}-\frac1{x^s}\right)~\mathrm dx$$
Which may be shown to equal
$$I'(s)=\begin{cases}\zeta(s)-1+\frac1{1-s}2^{1-s},&s\ne1\\\gamma,&s=1\end{cases}$$
And so,
\begin{align}I(1)&=\int_1^\infty\left(1-\zeta(s)-\frac{2^{1-s}}{1-s}\right)~\mathrm ds\\&=\gamma-1+\lim_{b\to\infty}\big(b-\zeta^{\star}(b)\big)-\lim_{a\to0^+}\big(\ln(-a\ln(2))-\zeta^\star(a+1)\big)\end{align}
where
$$\zeta^\star(s)=\int_t^s\zeta(x)~\mathrm dx,\quad t>1$$
and $\operatorname{Ei}(s)$ is the exponential integral.
...not really sure if this integral/limit is doable...
