Why is $\sum_{s=1}^{\infty }\left(-1\right)^{s+1} \left[\vphantom{\Large A}-2^{-s}\ \left(2^s-2\right)\ \zeta\left(s\right)\right]= 1-\log (2)$? could you give and explanation why 
$$
\sum_{s=1}^{\infty }\left(-1\right)^{s+1}
\left[\vphantom{\Large A}-2^{-s}\ \left(2^s-2\right)\ \zeta\left(s\right)\right]
=
1-\log (2)
$$ 
and $$\sum _{s=0}^{\infty } -2^{-2 s} \left(2^{2 s}-1\right) \zeta (2 s+1)=\frac{1}{4} \left[\psi ^{(0)}\left(\frac{i}{2}\right)+\psi ^{(0)}\left(-\frac{i}{2}\right)-\psi ^{(0)}\left(\frac{1}{2}+\frac{i}{2}\right)-\psi ^{(0)}\left(\frac{1}{2}-\frac{i}{2}\right)\vphantom{\LARGE A^{A}}\right]$$ 
i look in a forum and i think it is not convergent.
thanks
 A: The first formula, and in a similar way the second formula, most likely follows from the alternating zeta function (Dirichlet Eta function):
$$\eta(s):=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^s}=(1-2^{1-s})\zeta(s),$$
where the second equality is a fun exercise (hint: factor $1/k^s$ whenever it's even). Now, multiply both sides by $(-1)^{s+1}$ and sum both sides in $s$. The r.h.s. is exactly your sum in question, up to a minus sign. The sum in question looks like
$$\sum_{s=1}^\infty \sum_{k=1}^\infty\frac{(-1)^{k-1}(-1)^{s+1}}{k^s}$$
and as Steven rightfully pointed out, this is not absolutely convergent at $s=1$. However, as Daniel pointed out, the sum can actually be started at $s=2$, since on the r.h.s, $(1-2^{1-s})$ is 0 when $s=1$.  So the sum is really
$$\sum_{s=2}^\infty \sum_{k=1}^\infty \frac{(-1)^{k-1}(-1)^{s+1}}{k^{s}}$$
which now converges absolutely at $s=2$ and onward. So we can swap the double sums, and the important summation is then
$$\sum_{s=1}^\infty \frac{(-1)^{s+1}}{k^{s}}=\left(\frac{1}{k}-\frac{1}{k^2}+\frac{1}{k^3}-\cdots\right)=\left(-\frac{1}{1+1/k}+1\right)=\left(-\frac{k}{k+1}+1\right)=\frac{1}{k+1},$$
and now,
$$\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k+1}=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k+1}=\log(1+1)-1=\log(2)-1$$
since the l.h.s. is the Maclauren series of $\log(1+x)$ with $x=1$. 
