In my information theory course, we have been asked to find the entropy of a particular distribution in $\mathbb{N}$. In order to do so, I have come to the following integral $$\sum\limits_{n=1}^{\infty}\frac{\lfloor \log_2n \rfloor }{2^{2\lfloor \log_2n \rfloor}}$$ I would be content approximating it by $\sum\limits_{n=1}^{\infty}\dfrac{\log_2(n)}{ n^2}$, but I don't know how to compute it neither (or if this is even possible).

I know that the series $\sum\limits_{n=1}^{\infty}\dfrac{\log(n)}{ n^2}$ (with the natural logarithm for example), converges (Bertrand series), but I would like to know if its value is known.

I apologize if this has already been answered, or if it is easy to find somewhere else.

  • 2
    $\begingroup$ It's $-\zeta'(2)$, if that satisfies you. $\endgroup$ – Lord Shark the Unknown Mar 24 at 11:17
  • $\begingroup$ ... and $\zeta'(2)$ you will get with the first derivation of $\zeta(1-s)=\frac{2}{(2\pi)^s}\cos\frac{\pi s}{2}\Gamma(s)\zeta(s)$ at $s=2$ and $\zeta'(-1)=\frac{1}{2}-\ln A$ where $A$ is called the Glaisher-Kinkelin constant, e.g. here. But it's better you look at the answer of GEdgar. $\endgroup$ – user90369 Mar 24 at 11:49

With the integer part still in there $$ \sum\limits_{n=1}^{\infty}\dfrac{\lfloor \log_2n \rfloor }{2^{2\lfloor \log_2n \rfloor}} $$ proceed like this.

For $n=1$ we have $\lfloor \log_2n \rfloor = 0$.

For $2 \le n < 4$ we have $\lfloor \log_2n \rfloor = 1.\qquad$ (two terms)

For $4 \le n < 8$ we have $\lfloor \log_2n \rfloor = 2.\qquad$ (four terms)

And so on,

For $2^k \le n < 2^{k+1}$ we have $\lfloor \log_2n \rfloor = k.\qquad$ ($2^k$ terms)

So $$ \sum\limits_{n=1}^{\infty}\dfrac{\lfloor \log_2n \rfloor }{2^{2\lfloor \log_2n \rfloor}} = \sum_{k=0}^\infty 2^k\;\frac{k}{2^{2k}} = 2 $$

  • 1
    $\begingroup$ This has been incredibly helpful, much simpler than expected. Thank you! $\endgroup$ – Acas Mar 24 at 12:09

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.