Prove that $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left\lfloor\frac{\log(n)}{\log(2)}\right\rfloor=\gamma$

Prove that
$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left\lfloor\frac{\log(n)}{\log(2)}\right\rfloor=\gamma$

pProve that/p blockquote
p$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left\lfloor\frac{\log(n)}{\log(2)}\right\rfloor=\gamma$$/p
/blockquote pCan we find a known value for
$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left\lfloor\frac{\log(n)}{\log(k)}\right\rfloor$
for any $k\in\mathbb{N}?$/p