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

Martin Sleziak
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Dr. Wolfgang Hintze
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  • It turns out this is the same for $\sqrt[m]{k}$ for any positive integer $m$. See my generalization here: https://math.stackexchange.com/questions/3456631/if-m-in-mathbbn-then-sum-k-1-infty-dfrac-1k1-lfloor-sqrt – marty cohen Nov 30 '19 at 05:01
  • Dear downvoters, I would appreciate some information on your motivation to downvote my question. I suppose you took a responsible decision which means that you took into account what downvoting officially means, quote: "This question does not show any research effort. It is unclear and not useful." My comments: (1) In this case I had to conceil the solution steps in order not to spoil the simple and surprising result. Therefore I did not ask if there's a closed form but how it looks like. My research effort in closely related problems can be found in and via the reference I gave. – Dr. Wolfgang Hintze Dec 01 '19 at 20:02
  • Dear downvoters ctd. (2) The question is clearly indicated by bold letters in the text and the content should be clear (the type of question is common and abundant in this forum). (3) Whether the question is useful or not is to a high degreee a matter of taste. At least one upvoter, one generalizer and I considered it useful. Finally, I deplore missing consistency. My question was downvoted today 4 times by -2. Nevertheless this "bad" question has given the opportunity to gain of 7 upvotes in one answer (which I consider completely justified). – Dr. Wolfgang Hintze Dec 01 '19 at 20:03

2 Answers2

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

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

Riemann
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marty cohen
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  • @ Marty Cohen +1 Nicely done. Have you seen my last question https://math.stackexchange.com/q/3457629/198592 ? The minor change makes the problem much more complicated since, in your terms, the condition $n \in \mathbb{N} \implies f^{(-1)}(n)\in \mathbb{N} $ is voilated. I think it would be a real challenge to explore the limits of the methods we have seen so far. But there is "resistance" as well ... – Dr. Wolfgang Hintze Dec 01 '19 at 09:58
  • $u((n+1)^m-n^m-1)=1$ is obviously, because $(n+1)^m-n^m-1$ is even. (If $n$ is even, then $n+1$ is odd, or if $n$ is odd, then $n+1$ is even. In both case, the number $(n+1)^m-n^m-1$ is even.) – Riemann Jun 10 '20 at 08:33