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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.

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

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

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