# 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?

• Sorry, but the summation doesn't make sense. I fixed the math formulas, you should fix the limits of the summation. – egreg Apr 22 '15 at 9:21
• @marouane: I have edited your sum to reduce the confusion when the variable of summation is the same as a variable appearing outside the scope of the summation. Make sure it now reads as you intended. – robjohn Apr 22 '15 at 9:38
• Yes thanks this is what i need – marouane Apr 22 '15 at 9:40
• For the first approximation of the sum, you can replace the sum with an integral. – Daniel Fischer Apr 22 '15 at 9:43
• What is the bound of the sum in term of big O notation – marouane Apr 22 '15 at 9:45

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 :

• Just amazing thank y ou so much for time :) – marouane Apr 22 '15 at 13:45
• Hi marouane ! Did you see that your question has been noted "On hold" ? It will be deleted from the forum if you don't correct your question. Click on "edit" (at the bottom of your first post). Then add what it is asked : your thoughts on the problem and any attempts you have made to solve it . Type it in the main edition box (not in the box for comments). – JJacquelin Apr 22 '15 at 14:14
• The term for $k=1$ is $\frac{\log(0)}{0}$. The sum from $2$ to $100$ is $0.90750626$, which does not match your table. What am I missing? – robjohn May 2 '15 at 10:05
• Of course, there is a typo : the sum is from $k=5$ everywhere. – JJacquelin May 2 '15 at 10:30

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 $C\doteq-0.08334404437765197472024727705275296252855$.

• Thank you so much @rob for jour help :) – marouane Apr 22 '15 at 9:57
• Since the upper limit of the sum is $\lfloor\sqrt{n}\rfloor$, the error between $\sqrt{n}$ and $\lfloor\sqrt{n}\rfloor$ is between $0$ and $1$. The error due to this is $O\left(\frac{\log(\log(n))}{\sqrt{n}\log(n)}\right)$, so there is not much use in extending the asymptotic expansion further. – robjohn May 2 '15 at 10:03

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

Can you take it from here?

• Yes this help too , now i know how to calculate it thanks alot – marouane Apr 22 '15 at 10:04