<s>[INST] Bound $\sum_{k=1}^{n/2 } \frac{1}{ k \left( \log \frac{2n}{k}-\log(1+\frac{4}{k} \log \log\frac{2n}{k} ) \right)}$ I am looking to find an upper bound on
\begin{align}
\sum_{k=1}^{n/2 }   \frac{1}{  k \left( \log \frac{2n}{k}-\log(1+\frac{4}{k} \log \log\frac{2n}{k}  ) \right)}.
\end{align}
I am interested in the bound that is tight when $n$ is large.
I was thinking that for $k \in[1,n/2]$
\begin{align}
\log \frac{2n}{k}-\log(1+\frac{4}{k} \log \log\frac{2n}{k}  ) \ge  \log 4-\log(1+8 \log \log4  )=c,
\end{align}
where the minimum is attained at $k=n/2$ (I think this is true).
Then,
\begin{align}
\sum_{k=1}^{n/2 }   \frac{1}{  k \left( \log \frac{2n}{k}-\log(1+\frac{4}{k} \log \log\frac{2n}{k}  ) \right)} \le \frac{1}{c} \sum_{k=1}^{n/2 }   \frac{1}{  k} \le \frac{1}{c} (\log(n/2)+1).
\end{align}
Can one do better?
 [/INST] Start by bounding
\begin{align}
\sum_{k=1}^{n/2}   \frac{1}{  k \log \frac{2n}{k+4\log \log\frac{2n}{k}}} \leq \sum_{k=1}^{n/2}   \frac{1}{  k \log \frac{2n}{k+4\log \log2n}} \, .
\end{align}
Now for simplicity set $c=c(n)=4\log \log 2n$ and rearrange to $$\sum_{k=1}^{n/2} \frac{1}{k+c} \frac{1+\frac{c}{k}}{\log 2n - \log(k+c)} \, .$$
The second term is bounded by $$C\sum_{k=1}^{n/2} \frac{c}{k(k+c)} \leq C\sum_{k=1}^\infty \frac{c}{k(k+c)} \leq Cc\int_1^\infty\frac{{\rm d}k}{k(k+c)}=C\log(c+1)={\cal O}(\log \log \log n)$$ for some constant $C>0$.
The first term becomes $$\sum_{k=1}^{n/2} \frac{1}{k+c} \frac{1}{\log 2n - \log(k+c)} \leq \int_0^{n/2} \frac{{\rm d}k}{k+c}\frac{1}{\log 2n-\log(k+c)} \\
= \log \log \frac{2n}{c} - \log \log \frac{4n}{n+2c} = \log \log \frac{n}{2} + {\cal O}(\log \log\log \log n) \, .$$
$\log \log \frac{n/2}{\log n/2}$ is a good approximation.
</s>