I have been asked the following problem: Every real number $x$ can be written as a sum of the form $$ \sum_{i=1}^{\infty} \frac{a_n}{n^2}, $$ where $a_1\in \mathbb{Z}$, $a_2=0,1,2,3$, and $0\leq a_n<n$ are integers for $n\geq 3$.

(*)If $x$ is a rational number, then the expression can be of finitely many terms.

I don't know how to prove it. I have two expressions for $1/3$:

$$\frac{1}{3}=\frac{1}{2^2}+\frac{3}{6^2},$$ and $$\frac{1}{3}=\frac{1}{2^2}+\frac{1}{4^2}+\sum_{n=1}^{\infty}\frac{1}{49^n}.$$

I have no idea how to distinct them.

Could anyone provide me a proof/disproof of (*)?

Thank you every much!


2 Answers 2


Given $x>0$ a simple strategy is to choose, in the $n$-th step, the maximum $a_n$ such that $$ x-\sum_{j=1}^{n}\frac{a_j}{j^2}$$ is non-negative. Since the series $$\sum_{j=1}^{+\infty}\frac{j-1}{j^2}$$ is diverging, but the general term is infinitesimal, we can get arbitrarily close (i.e. converge towards) any positive real number. For example, let us represent $\frac{1}{3}$ this way. The first square bigger than $3$ is $4$, hence we just need to represent $\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$ as $\sum_{n\geq 3}\frac{a_n}{n^2}$. We have to choose $a_3=0$, since $\frac{1}{9}>\frac{1}{12}$, then by taking $a_4=1$ and just have to represent $\frac{1}{12}-\frac{1}{16}=\frac{1}{48}$ as $\sum_{n\geq 5}\frac{a_n}{n^2}$. This leads to your second formula (a representation with an infinite series).

However, as soon as we have to represent a number of the form $\frac{1}{mk^2}$ with $\gcd(m,k)=1$, we just have to subtract $\frac{m}{(mk)^2}$. So, in order to represent $x=\frac{p}{q}$ as a finite sum, we can first focus on decreasing the numerator. This is related to a famous problem.

Let us exhibit an algorithm for finding a representation of a rational number. By subtracting a few terms, we can assume to have to represent $\frac{p}{q}$ with $1<p<q$. The question is now: given $N$, it is possible to find a few terms of the form $\frac{m}{n^2}$, with $n>N,m<n$, such that $\frac{p}{q}-\sum\frac{m}{n^2}$ is positive and its numerator is one? The answer is affirmative. It is sufficient to take our $n$'s in the set of integers whose prime divisors belong to the prime divisors of $q$ and adjust, step by step, the base-$p_i$ representation of $\frac{p}{q}-\sum\frac{m}{n^2}$ for any prime divisor $p_i$ of $q$.

Another possible approach is just to show that the rational numbers that can be represented as sums of $\frac{m_i}{n_i^2}$ with $1\leq m_i<n_i$ are an additive semigroup of $\mathbb{Q}^+$. That is pretty straightforward to prove: we just have to sum two representations, then "fixing" terms like $\frac{l_i}{n_i^2}$ where $n_i\leq l_i\leq 2n_i-2$ by writing them as $\frac{n_i-1}{n_i^2}+\frac{4(l_i-n_i+1)}{(2n_i)^2}$. So we sum two representations and fix it by adjusting its terms, starting by the one with the lowest denominator. Since we can obviously represent $\frac{1}{n^2}$, we can represent $\frac{1}{n}$, hence any rational number.

  • $\begingroup$ Yes, if we got a good remaider $\frac{n}{mk^2}$ with $n<k$ such that $(mk)^2$ not being used as a denominater before, we are done at next step. But how do we get that? $\endgroup$
    – user119882
    Sep 20, 2014 at 20:23
  • $\begingroup$ @user119882: Thank you a lot, you just put me on the daily reputation leaderboard :D $\endgroup$ Sep 26, 2014 at 15:22

Just a comment:

Say $x=a/b$, with $a,b\in \mathbb{Z}$, $a,b>1$. Let $N$ be an integer such that $2^N>b$. Then let $a_0=[x]$, $a_j=[x\times 4^j]\%4$ for $j=1,2,\ldots, N$. Finaly, let $a_{N+1}=[x\times 4^N\times b^2]\%b^2$. Then one could check that $$ x=a_0+a_1\times 1/4+\cdots+a_N\times 1/4^N +a_{N+1}\times 1/(4^N\times b^2) $$ and all $a_i$ satisfy the required,

where, $m\%n$ denotes an integer number $y$ with $0\leq y\leq n-1$ such that $n|m-y$ and $[x]$ is the largest integer less or equal than $x$.


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