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