parmenides51 wrote: Prove that every rational number $x$ in the interval $(0, 1)$ can be written as a finite sum of different fractions of the type $\frac{1}{k(k + 1)}$ , that is, different elements in the sequence $\frac12$ , $\frac{1}{6}$ , $\frac{1}{12}$,$...$. Let $\frac pq$ (with positive integers $q>p>0$) be some rational number $\in(0,1)$ Let us build a sequence of positive integer quadruplets $(p_n,q_n,a_n,b_n)$ with $q_n>p_n>0$ in the following manner : 1) definition of the sequence : For $p_n,q_n$ : If $n=1$, $p_1=p$ and $q_1=q$ If $n>1$, $p_n=a_{n-1}b_{n-1}p_{n-1}-b_{n-1}q_{n-1}+a_{n-1}q_{n-1}$ and $q_n=a_{n-1}b_{n-1}q_{n-1}$ so that $\frac{p_n}{q_n}=\frac{p_{n-1}}{q_{n-1}}-\frac 1{a_{n-1}}+\frac 1{b_{n-1}}$ For $a_n,b_n$ : Let euclidian division $q_n=ap_n+r$ with $a\ge 1$ and $r\in\{0,1,,2,...,p_n-1\}$ If $r=0$, then $a\ge 2$ (since $q_n>p_n$) and define $a_n=a-1$ and $b_n=a(a-1)>a_n$ If $r\ne 0$, then define $a_n=a$ and $b_n=a+\left\lfloor\frac{p_na^2}r\right\rfloor>a_n$ (since $r<p_n$ and so $\left\lfloor\frac{p_na^2}r\right\rfloor\ge 1$ If $a_nb_np_n-b_nq_n+a_nq_n=0$ then sequence stops (it will be $p_{n+1}=0$) Else continue process 2) Claim : Sequence always eventually stops If $p_n|q_n$, then $a_nb_np_n-b_nq_n+a_nq_n=0$ and sequence stops If $p_n\not|q_n$ and $a_nb_np_n-b_nq_n+a_nq_n=0$ , then sequence stops too. Else $p_{n+1}=a_nb_np_n-b_nq_n+a_nq_n\ne 0$ Writing there $q_n=a_np_n+r$ and $b_n=a_n+\left\lfloor\frac{p_na_n^2}r\right\rfloor$, this becomes $p_{n+1}=r\left\{\frac{p_na_n^2}r\right\}<r\le p_n-1$ and so $p_{n+1}\le p_n-2$ And so $p_n$ is stictly decreasing and sequence must stop eventually. 3) Claim : $b_n<a_{n+1}$ Using again $q_n=ap_n+r$ (with $r\ne 0$ and $p_{n+1}\ne 0$ since otherwise sequence stops and $a_{n+1}$ does not exist) : If $a_n=a\ge 2$ (and so $b_n\ge 3$) : $q_{n+1}=a_nb_nq_n$ and $p_{n+1}=\le p_n-2$ and so : $a_{n+1}\ge \left\lfloor\frac{q_{n+1}}{p_{n+1}}\right\rfloor-1$ $>\frac{q_{n+1}}{p_{n+1}}-2$ $\ge \frac{a_nb_nq_n}{p_n-2}-2$ $>2b_n-2>b_n$ Q.E.D If $a_n=1$ (and so $b_n\ge 2$), then $q_n=p_n+r$ (with $p_n>r>0$) and $p_n=\alpha r+\beta$ (with $\alpha\ge 1$ and $\beta<r$) Then $b_n=1+\alpha$ and $p_{n+1}=\beta\ne 0$ (else sequence closed) and $q_{n+1}=b_nq_n=b_n((\alpha+1)r+\beta)\ge 3b_n\beta$ And $a_{n+1}\ge \left\lfloor\frac{q_{n+1}}{p_{n+1}}\right\rfloor-1$ $>\frac{q_{n+1}}{p_{n+1}}-2$ $\ge 3b_n-2>b_n$ Q.E.D 4) Required result : We have $\frac {p_{n+1}}{q_{n+1}}=\frac{p_n}{q_n}-\frac 1{a_n}+\frac 1{b_n}$ up to the end of sequence (when $p_{n+1}=0$) And so $\frac pq=\sum_{\text{over sequence}}\left(\frac 1{a_n}-\frac 1{b_n}\right)$ Which is also $\frac pq =\sum_{\text{over sequence}}\sum_{k=a_n}^{b_n-1}\frac 1{k(k+1)}$ And since $a_n<b_n<a_{n+1}$, all the $\frac 1{k(k+1)}$ are distinct, as required 5) some examples $\frac pq=\frac 7{111}$ $p_{1}=7\quad q_{1}=111\quad a_{1}=15\quad b_{1}=277$ $p_{2}=3\quad q_{2}=461205\quad a_{2}=153734\quad b_{2}=23634296490$ And so: $\frac{7}{111}=\frac 1{15}-\frac 1{277}+\frac 1{153734}-\frac 1{23634296490}$ $\frac{7}{111}=\sum_{k=15}^{276}\frac 1{k(k+1)}+\sum_{k=153734}^{23634296489}\frac 1{k(k+1)}$ $\frac pq=\frac{139}{211}$ $p_{1}=139\quad q_{1}=211\quad a_{1}=1\quad b_{1}=2$ $p_{2}=67\quad q_{2}=422\quad a_{2}=6\quad b_{2}=126$ $p_{3}=12\quad q_{3}=319032\quad a_{3}=26585\quad b_{3}=706788810$ And so : $\frac{139}{211}=\frac 1{1}-\frac 1{2}+\frac 1{6}-\frac 1{126}+\frac 1{26585}-\frac 1{706788810}$ $\frac{139}{211}=\sum_{k=1}^{1}\frac 1{k(k+1)}+\sum_{k=6}^{125}\frac 1{k(k+1)}+\sum_{k=26585}^{706788809}\frac 1{k(k+1)}$ $\frac pq=\frac{5}{7}$ $p_{1}=5\quad q_{1}=7\quad a_{1}=1\quad b_{1}=3$ $p_{2}=1\quad q_{2}=21\quad a_{2}=20\quad b_{2}=420$ And so : $\frac{5}{7}=\frac 1{1}-\frac 1{3}+\frac 1{20}-\frac 1{420}$ $\frac{5}{7}=\sum_{k=1}^{2}\frac 1{k(k+1)}+\sum_{k=20}^{419}\frac 1{k(k+1)}$