2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Sorry, but the summation doesn't make sense. I fixed the math formulas, you should fix the limits of the summation. $\endgroup$ – egreg Apr 22 '15 at 9:21
  • $\begingroup$ @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. $\endgroup$ – robjohn Apr 22 '15 at 9:38
  • $\begingroup$ Yes thanks this is what i need $\endgroup$ – marouane Apr 22 '15 at 9:40
  • $\begingroup$ For the first approximation of the sum, you can replace the sum with an integral. $\endgroup$ – Daniel Fischer Apr 22 '15 at 9:43
  • $\begingroup$ What is the bound of the sum in term of big O notation $\endgroup$ – marouane Apr 22 '15 at 9:45
4
$\begingroup$

On have to take care to the boundaries. One little picture says more than a long speech!

enter image description here $$ \sum_{k=5}^m\frac{\log(\log(k))}{k\log(k)} $$ $$\int_5^{m+1}\frac{\ln(\ln (x))}{x\ln(x)}dx<\sum_{k=5}^m\frac{\log(\log(k))}{k\log(k)}<\int_5^{m+1}\frac{\ln(\ln (x-1))}{(x-1)\ln(x-1)}dx$$ $$ \frac{1}{2}\left(\ln (\ln(m+1))\right)^2-\frac{1}{2}\left(\ln (\ln(5))\right)^2<\sum_{k=5}^m\frac{\log(\log(k))}{k\log(k)}<\frac{1}{2}\left(\ln (\ln(m))\right)^2-\frac{1}{2}\left(\ln (\ln(4))\right)^2 $$ 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 :

enter image description here

$\endgroup$
  • $\begingroup$ Just amazing thank y ou so much for time :) $\endgroup$ – marouane Apr 22 '15 at 13:45
  • $\begingroup$ 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). $\endgroup$ – JJacquelin Apr 22 '15 at 14:14
  • $\begingroup$ 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? $\endgroup$ – robjohn May 2 '15 at 10:05
  • $\begingroup$ Of course, there is a typo : the sum is from $k=5$ everywhere. $\endgroup$ – JJacquelin May 2 '15 at 10:30
4
$\begingroup$

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 $C\doteq-0.08334404437765197472024727705275296252855$.

$\endgroup$
  • $\begingroup$ Thank you so much @rob for jour help :) $\endgroup$ – marouane Apr 22 '15 at 9:57
  • $\begingroup$ 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. $\endgroup$ – robjohn May 2 '15 at 10:03
3
$\begingroup$

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

Can you take it from here?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.