Best approximation for $\displaystyle \sum_{k=2}^n\ln\ln k$? I need the best approximation for 
$\displaystyle{\sum_{k = 2}^{n}\ln\left(\ln\left(k\right)\right)}$. Any suggestion or hint is welcomed.
I derived $n\ln\left(\ln\left(n!\right) \over n\right)$ so is there any better one ?
 A: Euler-Maclaurin series:
$$ \sum_{k=2}^n \ln(\ln(k)) = C + \int_2^n \ln(\ln(t))\ dt + \dfrac{1}{2} \ln(\ln(n)) + \dfrac{1}{12 n \ln(n)} - \dfrac{1}{360 n^3 \ln(n)} - \dfrac{1}{240 n^3 \ln(n)^2} - \dfrac{1}{360 n^3 \ln(n)^3} + O(1/n^5) $$
for some constant $C$,
where $$\eqalign{\int_2^n \ln(\ln(t))\ dt &= n \ln(\ln(n)) - 2 \ln(\ln(2)) - \int_2^n \dfrac{dt}{\ln(t)}\cr &= n \ln(\ln(n)) - Li(n) - 2 \ln(\ln(2)) + Li(2)\cr}$$
Numerically it appears $C \approx -.2412388$.
A: $\newcommand{\+}{^{\dagger}}%
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With ${\it\mbox{Euler-Maclaurin Summation Formula}}$:
\begin{align}
&\sum_{k = 2}^{n}\ln\pars{\ln\pars{k}}
=\ln\pars{\ln\pars{2}} + \sum_{k = 1}^{n - 2}\ln\pars{\ln\pars{k + 2}}
\\[3mm]&\approx
\ln\pars{\ln\pars{2}} + \int_{0}^{n - 1}\ln\pars{\ln\pars{x + 2}}\,\dd x
-
{1 \over 2}\braces{\ln\pars{\ln\pars{0 + 2}}
+ \ln\pars{\ln\pars{\bracks{n - 1} + 2}}}
\\[3mm]&+ {1 \over 12}\braces{%
{1 \over \bracks{\pars{n - 1} + 2}\ln\pars{\bracks{n - 1} + 2}}
-
{1 \over \bracks{0 + 2}\ln\pars{0 + 2}}}
\\[3mm]&=
{1 \over 2}\,\ln\pars{\ln\pars{2}} - { 1 \over 24\ln\pars{2}} + \int_{0}^{n - 1}\ln\pars{\ln\pars{x + 2}}\,\dd x
-
{1 \over 2}\,\ln\pars{\ln\pars{n + 1}} + {1 \over 12\pars{n + 1}\ln\pars{n + 1}}
\end{align}
A: You can use the Euler-Maclaurin Sum Series. The first few terms are
$$
\int_2^n\log(\log(x))\,\mathrm{d}x+\frac12\log(\log(n))+C+\frac1{12n\log(n)}+\dots
$$
We won't worry about any terms past the first since we need to approximate the integral asymptotically, and the terms in that expansion are bigger than $\log(\log(n))$.
$$
\begin{align}
\int\log(\log(x))\,\mathrm{d}x
&=\int\log(u)\,\mathrm{d}e^u\\
&=e^u\log(u)-\int\frac{e^u}{u}\,\mathrm{d}u\\
&=e^u\log(u)-\frac{e^u}{u}-\int\frac{e^u}{u^2}\,\mathrm{d}u\\
&=e^u\log(u)-\frac{e^u}{u}-\frac{e^u}{u^2}-2\int\frac{e^u}{u^3}\,\mathrm{d}u\\
&=e^u\log(u)-\frac{e^u}{u}-\frac{e^u}{u^2}-2\frac{e^u}{u^3}-6\int\frac{e^u}{u^4}\,\mathrm{d}u\\
&=e^u\log(u)-\frac{e^u}{u}-\frac{e^u}{u^2}-2\frac{e^u}{u^3}-6\frac{e^u}{u^4}-24\int\frac{e^u}{u^5}\,\mathrm{d}u
\end{align}
$$
From this, we get the asymptotic expansion
$$
\sum_{k=2}^n\log(\log(k))\sim n\log(\log(n))-\frac{n}{\log(n)}\left(1+\frac1{\log(n)}+\frac2{\log(n)^2}+\frac6{\log(n)^3}+\dots\right)
$$
