Can we do that?

We note that:

$\begin{aligned} \sum_{n=1}^\infty \frac{(-1)^n}{n^s} + \sum_{n=1}^\infty \frac{1}{n^s} & = 2 \left(\frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{6^s} + ... \right) = 2 \sum_{n=1}^\infty \frac{1}{(2n)^s} = 2^{1-s} \sum_{n=1}^\infty \frac{1}{n^s} \\ \Rightarrow \sum_{n=1}^\infty \frac{(-1)^n}{n^s} & = 2^{1-s} \sum_{n=1}^\infty \frac{1}{n^s} - \sum_{n=1}^\infty \frac{1}{n^s} \\ & = \left(2^{1-s} - 1 \right) \color{#3D99F6}{\sum_{n=1}^\infty \frac{1}{n^s}} \end{aligned}$

We can rewrite $\begin{aligned} \color{#3D99F6}{\sum_{n=1}^\infty \frac{1}{n^s}} & = \frac{1}{1-s} \sum_{n=1}^\infty \left(\frac{n}{(n+1)^s} - \frac{n-s}{n^s} \right) \end{aligned}$ which converges for $\Re (s) > 0$ and it is found that $\zeta \left(\frac{1}{2} \right) \approx -1.4603545$ .

$\displaystyle \Rightarrow \sum_{n=1}^\infty \frac{(-1)^n}{n^s} = \left(2^{1-s} - 1 \right) \zeta \left(\frac{1}{2} \right)$

$\Rightarrow \log_2 (k+1) = \log_2 \sqrt{2} = \boxed{0.5}$

I don't know Cher-Seong Cheong sir did so here is my solution

First Note that

$\zeta \left( x \right) =\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ x } } }$

I would manipulate these terms by rearranging them ). We have

$\left( \frac { -1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ \sqrt { 3 } } +... \right) \\ =\left( \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 4 } } +\frac { 1 }{ \sqrt { 6 } } +... \right) -\left( \frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 3 } } +\frac { 1 }{ \sqrt { 5 } } +... \right) \\ =\left( \frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 3 } } +... \right) -2\left( \frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 3 } } +\frac { 1 }{ \sqrt { 5 } } +... \right) \\ =\zeta \left( 0.5 \right) -2\left( \frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 3 } } +... \right) +2\left( \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 4 } } +\frac { 1 }{ \sqrt { 6 } } +... \right) \\ =\zeta \left( 0.5 \right) -2\zeta \left( 0.5 \right) +\frac { 2 }{ \sqrt { 2 } } \left( \frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 3 } } +... \right) \\ =\left( \sqrt { 2 } -1 \right) \zeta \left( 0.5 \right)$

Hence $k=\sqrt(2)-1$

This implies $\log _{ 2 }{ \left( \sqrt { 2 } -1+1 \right) } =\log _{ 2 }{ \sqrt { 2 } } =0.5$

My solution will look similar to that of Comrade @Chew-Seong Cheong , but we need to think carefully about the domains of all the functions involved.

Consider the Dirichlet Eta Function $\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$ , defined when $\Re(s)>0$ . As Comrade Cheong observes, we have the equation $\eta(s)=(1-2^{1-s})\zeta(s)$ when $\Re(s)>1$ . Thus $\zeta(s)=(1-2^{1-s})^{-1}\eta(s)$ is the analytic continuation of $\zeta(s)$ when $\Re(s)>0$ and $2^{1-s}\neq 1$ . It follows that $k=2^{1-s}-1=2^{0.5}-1$ , and the answer is $\log_2(k+1)=\boxed{0.5}$ .