# $\sum _{n=0}^{\infty} \frac{1}{(n+1) (n+2)} \left(\frac{1}{\lfloor n \phi \rfloor +2}+\frac{1}{\lfloor n \phi ^{-1} \rfloor +2}\right)$

How to prove: $$\sum _{n=0}^{\infty} \frac{1}{(n+1) (n+2)} \left(\frac{1}{\lfloor n \phi \rfloor +2}+\frac{1}{\lfloor n \phi ^{-1} \rfloor +2}\right)=\frac{3}{4}$$ Here $$\phi=\frac{1+\sqrt 5}{2}$$ and $$\lfloor \cdot \rfloor$$ the floor function. I suspect this is related to number theory (continued fractions) which I'm not familiar with. Any help will be appreciated.

Update: Here is a related problem, solved by similar techniques.

• It may be relevant or not, but $\lfloor n \phi\rfloor$ is known as the “lower Wythoff sequence,” compare en.wikipedia.org/wiki/Beatty_sequence. Mar 16, 2020 at 8:23
• @pisco Would you mind letting me know which blog you found this identity at? I am interested in series involving floors of multiples of irrationals and I might like to take a look at the other identities on the blog. Mar 16, 2020 at 14:46
• @FranklinPezzutiDyer I have already copied most of the interesting posts on that blog. Mar 17, 2020 at 2:50

The term for $$n=0$$ gives us $$\frac 1 2$$.

For the others, use $$\frac 1{(n+1)(n+2)} = \frac 1 {n+1} - \frac 1{n+2}$$ break the sum into: $$\sum_{n=1}^\infty \frac 1{(n+1)(\lfloor n\phi\rfloor + 2)} - \sum_{n'=1}^\infty \frac 1 {(n'+2)(\lfloor n'\phi\rfloor + 2)} + \sum_{m'=1}^\infty \frac 1{(m'+1)(\lfloor m'\phi^{-1}\rfloor + 2)} - \sum_{m=1}^\infty \frac 1 {(m+2)(\lfloor m \phi^{-1}\rfloor + 2)}.$$ Note that each sum is absolutely convergent so there are no issues here. Next, use the following result.

Claim. For any integers $$n, n' \ge 1$$, we have $$m = \lfloor n\phi\rfloor \implies \lfloor m\phi^{-1}\rfloor = n-1, \qquad m' = \lfloor n'\phi\rfloor + 1 \implies \lfloor m'\phi^{-1}\rfloor = n'.$$ Proof. Since $$n\phi$$ is not an integer, $$m = \lfloor n\phi\rfloor$$ satisfies $$n\phi - 1 < m < n\phi$$. This gives $$n - \phi^{-1} < m \phi^{-1} < n$$ and thus $$\lfloor m\phi^{-1} \rfloor = n-1$$. The other case is similar.

Back to the problem. We see that every term in the first sum occurs in the fourth. Specifically, if $$m = \lfloor n \phi\rfloor$$ then $$(m+2)(\lfloor m\phi^{-1}\rfloor + 2) = (\lfloor n\phi\rfloor + 2)(n + 1)$$. Likewise, if $$m' = \lfloor n'\phi\rfloor + 1$$ then $$(m'+1)(\lfloor m'\phi^{-1}\rfloor + 2) = (\lfloor n'\phi\rfloor + 2)(n' + 2)$$. This results in a whole lot of cancellations, and the surviving terms are: $$- \sum_{m\in A} \frac 1 {(m+2)(\lfloor m\phi^{-1}\rfloor + 2)} + \sum_{m' \in B} \frac 1 {(m'+1)(\lfloor m'\phi^{-1}\rfloor + 2)},$$ where $$A$$ (resp. $$B$$) is the set of positive integers not expressible as $$\lfloor n\phi\rfloor$$ (resp. $$\lfloor n'\phi\rfloor + 1$$). Note that $$B = \{1\} \cup \{m+1 : m\in A\}$$. The case $$1\in B$$ gives us $$\frac 1 4$$. For the remaining terms, we claim that for all $$m\in A$$, we have $$\lfloor (m+1)\phi^{-1}\rfloor = \lfloor m\phi^{-1}\rfloor$$ which completes the proof since the two sums cancel each other out.

Claim. If positive integer $$m$$ is not expressible as $$\lfloor n\phi\rfloor$$, then $$\lfloor (m+1)\phi^{-1}\rfloor = \lfloor m\phi^{-1}\rfloor$$.

Proof. Since $$1 < \phi < 2$$, we have an integer $$n$$ such that $$\lfloor n \phi\rfloor = m-1, \qquad \lfloor (n+1)\phi \rfloor = m+1.$$ This gives the inequalities $$m-1 < n\phi < m$$ and $$m+1 < (n + 1) \phi < m+2$$ and thus $$n < m \phi^{-1} < n + 1 - \phi^{-1}$$. Since $$n+\phi^{-1} < (m+1)\phi^{-1} < n + 1$$ both floors are $$n$$.

• Very nice solution! The proof of last claim can be simplified into: if $n=\lfloor (m+1)\phi^{-1}\rfloor > \lfloor m\phi^{-1}\rfloor$, then $(m+1)\phi^{-1} > n > m\phi^{-1}$, so $m=\lfloor n\phi \rfloor$, contradiction to choice of $m$. This removes the assumption $1<\phi<2$. By replacing $\phi$ with any positive irrational, the sum is still $3/4$. Mar 16, 2020 at 13:32