Keep it simple

$\zeta=\sum_{n=1}^\infty \dfrac{1}{n^{2.5}}\\ S=\sum_{n=1}^\infty \dfrac{(-1)^n}{n^{2.5}}=-\dfrac{1}{1}+\dfrac{1}{2^{2.5}}-\dfrac{1}{3^{2.5}}+\dfrac{1}{4^{2.5}}\pm......\\ S=-\left(\dfrac{1}{1}+\dfrac{1}{2^{2.5}}+\dfrac{1}{3^{2.5}}+\dfrac{1}{4^{2.5}}+....\right)+2\left(\dfrac{1}{2^{2.5}}+\dfrac{1}{4^{2.5}}+.....\right)\\ S=-\left(\dfrac{1}{1}+\dfrac{1}{2^{2.5}}+\dfrac{1}{3^{2.5}}+\dfrac{1}{4^{2.5}}+....\right)+\dfrac{2}{2^{2.5}}\left(\dfrac{1}{1}+\dfrac{1}{2^{2.5}}+\dfrac{1}{3^{2.5}}+\dfrac{1}{4^{2.5}}+.....\right)\\ S=-\zeta+2^{1-2.5}\zeta=(2^{-1.5}-1)\zeta\\ \log_{2} (2^{-1.5}-1+1)=\boxed{-1.5}$

$\begin{aligned} \sum_{n=1}^\infty \frac {(-1)^n}{n^{2.5}} & = - \sum_{k=1}^\infty \frac 1{n^{2.5}} + 2 \sum_{k=1}^\infty \frac 1{(2n)^{2.5}} \\ & = - \sum_{k=1}^\infty \frac 1{n^{2.5}} + \frac 2{2^{2.5}} \sum_{k=1}^\infty \frac 1{n^{2.5}} \\ & = \left(2^{-1.5}-1\right) \sum_{k=1}^\infty \frac 1{n^{2.5}} \end{aligned}$

Therefore, $k= 2^{-1.5}-1$ and $\log_2 (k+1) = \boxed{-1.5}$ .

Let's define $A = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{2.5}}, B = \sum_{n=1}^{\infty} \frac{1}{n^{2.5}}$

Then $k = \frac{A}{B}$ . But this would not be an easy computation to do directly. However, notice that $A + B = \sum_{n=1}^{\infty} \frac{2}{(2n)^{2.5}} = 2^{-1.5}B$ . Divide by $B$ to obtain $\frac{A}{B} + 1 = 2^{-1.5}$ . The solution is then $-1.5$ .