Closed form of sum $\sum\limits_{k=1}^{\infty } \frac{(-1)^{k+1}}{\left\lfloor \sqrt{k}\right\rfloor}$ In two previous problems (Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$ and Closed expression for sum $\sum_{k = 1}^{\infty} \frac{\left\lfloor \sqrt{k} \right \rfloor}{k^2}$) the infinite sums contained the floor function (of a square root) in the numerator. 
Here we ask, in a simple example, what happens if the floor function is in the denominator.
Question: what is the closed form of $\sum _{k=1}^{\infty } \frac{(-1)^{k+1}}{\left\lfloor \sqrt{k}\right\rfloor }$
 A: OK. Here's my generalization
which ran into some resistance
when posted as a separate problem.
Let $f(x)$ be such that
$f(1) = 1,
f'(x) > 0, f''(x) < 0,
f(x) \to \infty,
n \in \mathbb{N} 
\implies f^{(-1)}(n)\in \mathbb{N}
$.
($f^{(-1)}(n)$
is the inverse function of $f$)
What can we say about
$$S=\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{\lfloor f(k) \rfloor} $$
Let $g$ be the
inverse function of $f$,
so
$f(g(x)) = g(f(x)) = x
$.
Let 
$u(n) =
\begin{cases}
0 \text{ if } n \text{ odd}\\
1 \text{ if } n \text{ even}\\
\end{cases}
=\dfrac{(-1)^n+1}{2}.
$
\begin{align}
S
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{\lfloor f(k) \rfloor}\\
&=\sum_{n=1}^{\infty} \sum_{k=g(n)}^{g(n+1)-1} \dfrac{(-1)^{k+1}}{\lfloor f(k) \rfloor}\\
&=\sum_{n=1}^{\infty} \sum_{k=g(n)}^{g(n+1)-1} \dfrac{(-1)^{k+1}}{\lfloor n \rfloor}\\
&=\sum_{n=1}^{\infty} \dfrac1{n}\sum_{k=g(n)}^{g(n+1)-1} (-1)^{k+1}\\
&=\sum_{n=1}^{\infty} \dfrac1{n}\sum_{k=0}^{g(n+1)-g(n)-1} (-1)^{k+g(n)+1}\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{g(n)+1}}{n}\sum_{k=0}^{g(n+1)-g(n)-1} (-1)^{k}\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{g(n)+1}}{n}u(g(n+1)-g(n)-1)\\
\end{align}
If $f(k) = \sqrt{k}$,
then
$g(n) = n^2$
so
$u(g(n+1)-g(n)-1)
=u(2n)
=1
$
and
$(-1)^{g(n)+1}
=(-1)^{n^2+1}
=(-1)^{n+1}
$
so
$$S
=\sum_{n=1}^{\infty} \dfrac{(-1)^{g(n)+1}}{n}u(g(n+1)-g(n)-1)
=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}
=\ln(2).
$$
If $f(k) = \sqrt[3]{k}$,
then
$g(n) = n^3$
so
$u(g(n+1)-g(n)-1)
=u(3n^2+3n)
=u(3n(n+1))
=1
$
and
$(-1)^{g(n)+1}
=(-1)^{n^3+1}
=(-1)^{n+1}
$
so
$$S
=\sum_{n=1}^{\infty} \dfrac{(-1)^{g(n)+1}}{n}u(g(n+1)-g(n)-1)
=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}
=\ln(2).
$$
If $f(k) = \sqrt[m]{k}$,
then
$g(n) = n^m$
so
$\begin{array}\\
u(g(n+1)-g(n)-1)
&=u((n+1)^m-n^m-1)\\
&=u(\sum_{j=1}^{m-1} \binom{m}{j}n^j)\\
&=u(\sum_{j=1}^{\lfloor \frac{m-1}{2} \rfloor} (\binom{m}{j}n^j+\binom{m}{m-j}n^{m-j})
\qquad\text{central binomial coefficient is even}\\
&=u(\sum_{j=1}^{\lfloor \frac{m-1}{2} \rfloor} (\binom{m}{j}(n^j+n^{m-j}))\\
&=u(\sum_{j=1}^{\lfloor \frac{m-1}{2} \rfloor} (\binom{m}{j}n^j(1+n^{m-2j}))\\
&=1
\qquad\text{since }n^j(1+n^{m-2j})
\text{ is even}\\
\end{array}
$
and
$(-1)^{g(n)+1}
=(-1)^{n^m+1}
=(-1)^{n+1}
$
so
$$S
=\sum_{n=1}^{\infty} \dfrac{(-1)^{g(n)+1}}{n}u(g(n+1)-g(n)-1)
=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}
=\ln(2)
$$
A: For any $k \geq 1$, we can uniquely write $k=n^2+\ell$, with $0\leq \ell \leq 2n$. Then, $\lfloor \sqrt{k}\rfloor = n$, so that
$$\begin{align}
\sum_{k=1}^\infty \frac{(-1)^{k+1}}{\lfloor \sqrt{k}\rfloor}
&= \sum_{n=1}^\infty\sum_{\ell=0}^{2n} \frac{(-1)^{n^2+\ell+1}}{n}
= -\sum_{n=1}^\infty \frac{(-1)^{n^2}}{n} \sum_{\ell=0}^{2n} (-1)^{\ell}\\
&= -\sum_{n=1}^\infty \frac{(-1)^{n}}{n} \sum_{\ell=0}^{2n} (-1)^{\ell}
= -\sum_{n=1}^\infty \frac{(-1)^{n}}{n}
\end{align}$$
since $(-1)^{n^2}=(-1)^n$ and $\sum_{\ell=0}^{2n} (-1)^{\ell}=1$ for all $n$.
It follows that
$$
\sum_{k=1}^\infty \frac{(-1)^{k+1}}{\lfloor \sqrt{k}\rfloor}
= -\sum_{n=1}^\infty \frac{(-1)^{n}}{n} = \boxed{\log 2}
$$
