Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$ Inspired by the recent question if the series $\sum_{k=1}^{\infty} \frac{\sqrt{k}-\left\lfloor \sqrt{k}\right\rfloor}{k}$ diverges (which is the case) I became interested in the alternating series which is convergent by the Leibniz criterion. 
The core of the problem is then the question if this sum
$$s = \sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}\simeq 0.591561$$
has a closed expression. Here $\left\lfloor {x}\right\rfloor$ is the greatest integer less than or equal to $x$.
I have found a nice integral representation for $s$ but I could not find a closed expression. Also, due to the slow convergence of the sum it is not trivial to get a numerical result with high accuracy which might be necessary to identify a possible closed expression.
Problems
a) find a closed expression for $s$
b) find the numerical result exact to 20 decimal places
 A: Update
We may use the convergence acceleration of alternating series developed by Cohen, Villegas, and Zagier.
Let
$$s = \ln 2 + \sum_{n=1}^\infty (-1)^n n
\sum_{i=1}^{n} \frac{1}{(n^2 + 2i-1)(n^2+2i)}$$
and
$$s_n = \ln 2 + \sum_{k=1}^n \frac{c_{n,k}}{d_n}\sum_{i=1}^k  \frac{k}{(k^2 + 2i-1)(k^2+2i)}$$
where
\begin{align}
d_n &= \frac{(3+\sqrt{8})^n + (3-\sqrt{8})^n}{2}, \\
c_{n,k} &= (-1)^k \sum_{m=k+1}^n \frac{n}{n+m} \binom{n+m}{2m} 2^{2m}.
\end{align}
From Proposition 1 in [1], we have
$$|s-s_n| \le \frac{s}{d_n}.$$
Maple: $\mathrm{evalf}(s, 30) = 0.591560779349817340213846903345$,
$\mathrm{evalf}(s_{28} - s, 30) = 1.6944769437\cdot 10^{-21}.$
[1] Henri Cohen, Fernando Rodriguez Villegas, and Don Zagier,
"Convergence Acceleration of Alternating Series".
Previously written
We have
\begin{align}
s &= \sum_{k=1} (-1)^{k+1} \frac{\lfloor \sqrt{k}\rfloor}{k}\\
&= \sum_{n=1}^\infty \left(\sum_{k = n^2} (-1)^{k+1} \frac{\lfloor \sqrt{k}\rfloor}{k}\right)
+ \sum_{n=1}^\infty \left(\sum_{n^2 < k < (n+1)^2} (-1)^{k+1} \frac{\lfloor \sqrt{k}\rfloor}{k}\right)\\
&= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} - \sum_{n=1}^\infty (-1)^n n 
\sum_{i=1}^{2n} (-1)^i \frac{1}{n^2 + i}\\
&= \ln 2 - \sum_{n=1}^\infty (-1)^n n
\sum_{i=1}^{2n} (-1)^{i} \frac{1}{n^2 + i}\\
&= \ln 2 + \sum_{n=1}^\infty (-1)^n n
\sum_{i=1}^{n} \frac{1}{(n^2 + 2i-1)(n^2+2i)}.\tag{1}
\end{align}
Maple can give numerical approximation of (1) with high accuracy. Or we may use "Convergence Acceleration of Alternating Series" technique to calculate (1). 
