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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Martin R
    Mar 16, 2020 at 8:23
  • $\begingroup$ @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. $\endgroup$ Mar 16, 2020 at 14:46
  • $\begingroup$ @FranklinPezzutiDyer I have already copied most of the interesting posts on that blog. $\endgroup$ Mar 17, 2020 at 2:50

1 Answer 1


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 $n$.

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    $\begingroup$ 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 $3/4$. $\endgroup$
    – pisco
    Mar 16, 2020 at 13:32

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