Closed form of sum $\sum\limits_{k=1}^{\infty } \frac{(-1)^{k+1}}{\left\lfloor \sqrt{k}\right\rfloor}$

Question

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 $\sum _{k=1}^{\infty } \frac{(-1)^{k+1}}{\left\lfloor \sqrt{k}\right\rfloor }$

Answer 1

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 $$ \sum_{k=1}^\infty \frac{(-1)^{k+1}}{\lfloor \sqrt{k}\rfloor} = -\sum_{n=1}^\infty \frac{(-1)^{n}}{n} = \boxed{\log 2} $$

Answer 2

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 $$S =\sum_{n=1}^{\infty} \dfrac{(-1)^{g(n)+1}}{n}u(g(n+1)-g(n)-1) =\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n} =\ln(2) $$