Approximation of a sum $\sum ^{\sqrt{n} }_{k=5}\frac{\log\log(k)}{k\log(k)} $

Question

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?

Answer 1

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 $C\doteq-0.08334404437765197472024727705275296252855$.

Answer 2

$$\sum_{k=5}^{\sqrt n}\frac{\ln(\ln k)}{k\ln k}\approx \int\limits_5^{\sqrt n}\frac{\ln(\ln k)}{k\ln k}\,\mathrm dk\stackrel{t=\ln k}=\int\limits_{\ln 5}^{0.5\ln n}\frac{\ln t}{t}\,\mathrm dt\stackrel{u=\ln t}=\int\limits_{\ln(\ln 5)}^{\ln(0.5\ln n)}u\,\mathrm du$$

Can you take it from here?

Answer 3

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 :