Infinite Series $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n}\left\lfloor\frac{\log(n)}{\log(2)}\right\rfloor$ How to prove that
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left\lfloor\frac{\log(n)}{\log(2)}\right\rfloor=\gamma$$
Can 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 $k\in\mathbb{N}?$
 A: First show that the sum converges by grouping summands in pairs, so it is enough to check the approximating sums with $n$ running from $1$ to $2^k-1$, say. For these, split the sum into the pieces where $\left\lfloor\frac{\log n}{\log 2}\right\rfloor$ is constant: $$\sum_{n=1}^{2^k-1} \frac{(-1)^n}{n} \left\lfloor \frac{\log n}{\log 2} \right\rfloor = \sum_{r=0}^{k-1} \sum_{n=2^r}^{2^{r+1}-1} r\frac{(-1)^n}{n} = \sum_{r=0}^{k-1} r \sum_{n=2^r}^{2^{r+1}-1} \frac{(-1)^n}{n} = \sum_{r=1}^{k-1} r \sum_{n=2^r}^{2^{r+1}-1} \frac{(-1)^n}{n}$$
Split $\frac{1}{n}$ for the even n's into $\frac{1}{n/2}-\frac{1}{n}$ and rearrange the inner sums:
$$\begin{align}\sum_{r=1}^{k-1} r \sum_{n=2^r}^{2^{r+1}-1} \frac{(-1)^n}{n}&= \sum_{r=1}^{k-1} r \left( \sum_{n=2^r}^{2^{r+1}-1} \frac{-1}{n} + \sum_{n=2^{r-1}}^{2^r-1} \frac{1}{n}\right)\\&
=1+\sum_{r=1}^{k-2}(r+1-r) \sum_{n=2^r}^{2^{r+1}-1} \frac{1}{n} - (k-1)\sum_{n=2^{k-1}}^{2^k-1}\frac{1}{n}\\&= \sum_{n=1}^{2^{k-1}-1} \frac{1}{n} - (k-1)\sum_{n=2^{k-1}}^{2^k-1}\frac{1}{n}.\end{align}$$
The first sum is $\log 2^{k-1} + \gamma + o(1)$, the second is $(k-1) \log 2 + o(1)$.
For $k$ instead of $2$, you can do the same rearranging, but the sums no longer overlap, so it's trickier to handle.
