TITLE: Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$
QUESTION [11 upvotes]: 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

REPLY [2 votes]: Partial solution and alternative characterisation
We may rephrase the problem with the following rearrangement, reducing the problem to the double-series in red, which I will refer to as $d$.
$$\begin{align} s
&=\sum_{n\ \text{square}}\frac{\left(-1\right)^{n+1}}{n}\lfloor\sqrt{n}\rfloor
+\sum_{n\ \text{not square}}\frac{\left(-1\right)^{n+1}}{n}\lfloor\sqrt{n}\rfloor
\tag{1}
\\
&=-\sum_{n\ \text{square}}\frac{\left(-1\right)^{n}}{n}\sqrt{n}
-\sum_{n\ \text{not square}}\frac{\left(-1\right)^{n}}{n}\left(\sqrt{n}-\tfrac12+\tfrac1\pi\sum_{k\ge 1}\frac{\sin(2\pi k\sqrt{n})}{k}\right)
\tag{2}
\\
&=-\left[\sum_{n\ \text{square}}\frac{\left(-1\right)^{n}}{\sqrt{n}}
+\sum_{n\ \text{not square}}\frac{\left(-1\right)^{n}}{\sqrt{n}}\right]\ldots
\\&\quad\ldots+\tfrac12\color{blue}{\sum_{n\ \text{not square}}\frac{\left(-1\right)^{n}}{n}}-\tfrac1\pi\sum_{n\ \text{not square}}\sum_{k\ge 1}\frac{\left(-1\right)^{n}\sin(2\pi k \sqrt{n})}{nk}
\\
\\
&=-\left(\sqrt{2}-1\right)\zeta\left(\tfrac12\right)+\frac12\left(\frac{{\pi^{2}}}{12}-{\ln 2}\right)-\frac{1}{\pi}\color{red}{\sum_{n\geq 1}\sum_{k\geq 1}\frac{\left(-1\right)^{n}\sin\left(2\pi k\sqrt{n}\right)}{nk}}
\tag{3}
\end{align}$$
In $(1)$, we separate the sum for $s$ into two parts, one over the squares and one over all other naturals, to account for the discontinuity of the following Fourier series at the integers. In $(2)$, we use the Fourier series of the floor function, $\lfloor x\rfloor=\displaystyle x-\tfrac12+\tfrac1\pi\sum_{k\ge 1}\frac{\sin(2\pi kx)}{k}$. In $(3)$, we use that $\sum_{k\geq1}\frac{(-1)^k}{\sqrt{k}}=(\sqrt2-1)\zeta(\frac12)$ (Question 2059712) and that $\sin(2\pi n)=0$ for $n\in\mathbb{N}$. We may evaluate the series in blue with
$$\begin{aligned}
\sum_{n\ \text{not square}}\frac{\left(-1\right)^{n}}{n}
&=\sum_{n\ge1}\frac{\left(-1\right)^{n}}{n}-\sum_{n\ \text{square}}\frac{\left(-1\right)^{n}}{n}
\\
&=-\ln2-\sum_{k\ge1}\frac{\left(-1\right)^{k^2}}{k^2}
\\
&=-\ln2-\frac{-\pi^2}{12}
\end{aligned}
$$
by using $\sum_{k\geq1}\frac{(-1)^k}{k^2}=\frac{-\pi^2}{12}$ and that $k^2$ has the same parity as $k$.

Partial analytic solution to $f_1(k)$
The two sums over $n$ and $k$, in the series for $d$, can be swapped, which gives us $\displaystyle d=\frac{-1}{\pi}\sum_{k\ge1}\frac{f_1(k)}{k}$, where $\displaystyle f_1(k)=\sum_{n\geq1}\frac{(-1)^n\sin(2\pi k\sqrt{n})}{n}$, which has the fairly simple Laplace transform $\frac{1}{2\pi}F\left(\frac{x}{2\pi}\right)$, where $\displaystyle F(x)=\sum_{n\ge 1}\frac{\left(-1\right)^{n}}{\sqrt{n}\left(n+x^2\right)}$. Therefore, $\displaystyle d=-\frac{1}{2\pi^{2}}\sum_{k=1}^{\infty}\frac{\mathcal{L}^{-1}\left\{F\left(\frac{s}{2\pi}\right),k\right\}}{k}$.
$F(x)$ has a similar series to the zeta functions. Substituting $x$ for $x^2$ in $F$ and taking the $k$'th derivative gives us $\small\displaystyle \frac{\mathrm{d}^k}{\mathrm{d}x^k}\sum_{n\ge 1}\frac{\left(-1\right)^{n}}{\sqrt{n}\left(n+x\right)}= \sum_{n\ge 1}\frac{(-1)^{k}k!\left(-1\right)^{n}}{\sqrt{n}\left(n+x\right)^{k+1}}$, so we have the interesting Maclaurin series, $\small \displaystyle F(x)=\sum_{k\ge0}\left(-1\right)^{k+1}\left[1-{2^{-\frac{1}{2}-k}}\right]\zeta\left(k+\frac{3}{2}\right)x^{2k}$. Although this only converges within a small radius of convergence, it does hint at the possibility of a closed form or analytic extension of this series.

Comments on further directions
The expression for $s$ converges slower than River Li's but could possibly be tractable. It is similar to the series evaluated in the following questions, which would suggest that it could be amenable to the Euler-Maclaurin, Abel-Plana or Poisson summation formulae. It is also possible that it could be expressed in terms of theta or Bessel functions, though I am yet to find whether this is the case.

Question 3374399: $\sum_{n\geq1}\sin(\pi\sqrt{n^2+1})$

Question 3364125: $\sum_{n\geq1}\frac{\sin\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}$

Question 27123 : $\sum_{n\ge1}\frac{\sin\left(\sqrt{n}x\right)}{n}$

Question 3352737: $\sum_{1\leq n \leq x}\sin\sqrt{n}$

Question 2270: $\sum_{n\geq 1}\frac{\sin(n^x)}{n}$



Addressing conflicting evaluation of $\displaystyle\sum_k$
Wolfgang's analytic solution in the comments for the sum over $k$ had an unexpected value, which I believe was due to $\displaystyle
\sum_{k\ge1}\frac{\sin(2\pi k \sqrt{n})}{k}
=
\frac{i\operatorname{Log}\left({-e^{-2\pi i\sqrt{n}}}\right)}{2}$ only being true for $\sqrt{n}\notin\mathbb{Z}$, where $\operatorname{Log}(z)$ is the principal branch of the complex logarithm. This is because, when $m\in\mathbb{Z}$, we have $\sin(2\pi m)\equiv0$ but $\operatorname{Log}(-e^{-2\pi i m})\equiv\operatorname{Log}(-1)\equiv \pi i$. This is effectively the same problem that made me remove the squares from the earlier Fourier series. If you take into account the discrepant sum over $\sqrt{n}\in\mathbb{Z}$, I believe the two series should be equal.
$$\begin{align}d&=\frac{-i}{2\pi}\sum_{n\ge1}\frac{(-1)^n\operatorname{Log}(-e^{2\pi i \sqrt{n}})}{n}-\left(\sum_{m\ge 1}\frac{-i(-1)^{m^2}(\pi i)}{2\pi m^2}\right)
\\
&=\underbrace{-0.489\ldots}_{\approx -0.336072}-\frac12\sum_{m\ge 1}\frac{(-1)^{m}}{m^2}
\\
&=-0.489-\frac12\left(\frac{-\pi^2}{12}\right)
\\
&=-0.077
\end{align}$$