Approximation of a sum $\sum ^{\sqrt{n} }_{k=5}\frac{\log\log(k)}{k\log(k)} $ What method could I use to obtain an approximation of this sum
$$\sum^{\sqrt{n}}_{k=5}\frac{\log\log(k)}{k\log(k)}$$
Should I proceed by an integral? How can I calculate its lower and upper bound? 
 A: On have to take care to the boundaries. One little picture says more than a long speech!

$$
\sum_{k=5}^m\frac{\log(\log(k))}{k\log(k)}
$$
$$\int_5^{m+1}\frac{\ln(\ln (x))}{x\ln(x)}dx<\sum_{k=5}^m\frac{\log(\log(k))}{k\log(k)}<\int_5^{m+1}\frac{\ln(\ln (x-1))}{(x-1)\ln(x-1)}dx$$
$$
\frac{1}{2}\left(\ln (\ln(m+1))\right)^2-\frac{1}{2}\left(\ln (\ln(5))\right)^2<\sum_{k=5}^m\frac{\log(\log(k))}{k\log(k)}<\frac{1}{2}\left(\ln (\ln(m))\right)^2-\frac{1}{2}\left(\ln (\ln(4))\right)^2
$$
The mean value is a very good approximate :
$$\sum_{k=5}^m\frac{\log(\log(k))}{k\log(k)}\simeq \frac{1}{4}\left(\left(\ln (\ln(m+1))\right)^2-\left(\ln (\ln(5))\right)^2 + \left(\ln (\ln(m))\right)^2-\left(\ln (\ln(4))\right)^2\right)$$
An even better approximate is obtained in considering the integral of the "mean" function $y=\int_5^{m+1}\frac{\ln(\ln (x-0.5))}{(x-0.5)\ln(x-0.5)}dx$
$$\sum_{k=5}^m\frac{\log(\log(k))}{k\log(k)}\simeq \int_5^{m+1}\frac{\ln(\ln (x-0.5))}{(x-0.5)\ln(x-0.5)}dx=\int_{4.5}^{m+0.5}\frac{\ln(\ln (x))}{x\ln(x)}dx=\frac{1}{2}\left(\ln (\ln(m+0.5))\right)^2-\frac{1}{2}\left(\ln (\ln(4.5))\right)^2$$
The comparison is shown below :

A: The Euler-Maclaurin Summation Formula says
$$
\begin{align}
\sum_{k=5}^m\frac{\log(\log(k))}{k\log(k)}
&=\frac12\log(\log(m))^2+C+O\left(\frac{\log(\log(m))}{m\log(m)}\right)
\end{align}
$$
Therefore,
$$
\begin{align}
&\sum_{k=5}^{\sqrt{n}}\frac{\log(\log(k))}{k\log(k)}\\
&=\tfrac12\log(\log(\sqrt{n}))^2+C+O\left(\frac{\log(\log(n))}{\sqrt{n}\log(n)}\right)\\
&=\tfrac12\log\left(\tfrac12\log(n)\right)^2+C+O\left(\frac{\log(\log(n))}{\sqrt{n}\log(n)}\right)\\
&=\tfrac12\log(\log(n))^2-\log(2)\log(\log(n))+\tfrac12\log(2)^2+C+O\left(\frac{\log(\log(n))}{\sqrt{n}\log(n)}\right)
\end{align}
$$
where $C\doteq-0.08334404437765197472024727705275296252855$.
A: $$\sum_{k=5}^{\sqrt n}\frac{\ln(\ln k)}{k\ln k}\approx \int\limits_5^{\sqrt n}\frac{\ln(\ln k)}{k\ln k}\,\mathrm dk\stackrel{t=\ln k}=\int\limits_{\ln 5}^{0.5\ln n}\frac{\ln t}{t}\,\mathrm dt\stackrel{u=\ln t}=\int\limits_{\ln(\ln 5)}^{\ln(0.5\ln n)}u\,\mathrm du$$
Can you take it from here?
