How would one go to prove that
$$\gamma = \frac{1}{2} + \sum_{n=1}^{\infty} \frac{\left \lfloor \log_2 n \right \rfloor}{2n(2n+1)(2n+2)}$$
where $\gamma$ stands for the Euler - Mascheroni constant and $\lfloor \cdot \rfloor$ is the floor function?
Honestly , I have no idea how to tackle it.
Simple creative telescoping sums reveal $$\sum_{n=1}^{\infty} \frac{\left \lfloor \log_2 n \right \rfloor}{2n(2n+1)(2n+2)}=\sum_{k=0}^{\infty} \sum_{n=2^k}^{2^{k+1}-1} \frac{\left \lfloor \log_2 n \right \rfloor}{2n(2n+1)(2n+2)}$$ $$=\sum_{k=0}^{\infty} k\sum_{n=2^k}^{2^{k+1}-1} \frac{1}{2n(2n+1)(2n+2)}$$ $$=-\frac{1}{4}+\lim_{n\to\infty}(\underbrace{\sum_{k=0}^n\left((k+1) \psi(2^{k+1})-k\psi(2^{k})\right)}_{\text{telescoping sum}}+\underbrace{\sum_{k=0}^n\left(k\psi(2^{k+1})-(k+1)\psi(2^{k+2})\right)}_{\text{telescoping sum}}$$ $$+\underbrace{\sum_{k=0}^n(\psi(2^{k+2})-\psi(2^{k+1}))}_{\text{telescoping sum}})$$ $$=\gamma+\log(2)-5/4.$$
Q.E.D.
Reference: https://en.wikipedia.org/wiki/Digamma_function.