# Is it known the value of $\sum\limits_{n=1}^{\infty}\frac{\log(n)}{ n^2}$?

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.

• It's $-\zeta'(2)$, if that satisfies you. – Lord Shark the Unknown Mar 24 at 11:17
• ... 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. – 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 $$\sum\limits_{n=1}^{\infty}\dfrac{\lfloor \log_2n \rfloor }{2^{2\lfloor \log_2n \rfloor}} = \sum_{k=0}^\infty 2^k\;\frac{k}{2^{2k}} = 2$$

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