MHB How Do Disjoint Intervals Solve the Inequality in POTW #409?

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The discussion focuses on solving the inequality involving the sum of fractions, specifically showing that the set of real numbers satisfying the inequality forms a union of disjoint intervals. The problem states that the total length of these intervals is 1988. The deadline for submissions has been extended to encourage more participation from members. There has been a lack of responses to the problem, prompting the extension. The thread emphasizes the importance of member engagement in finding solutions.
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Here is this week's POTW:

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Show that the set of real numbers $x$ which satisfy the inequality $$\sum_{k=1}^{70}\dfrac{k}{x-k}\ge \dfrac{5}{4}$$ is a union of disjoint intervals, the sum of whose lengths is 1988.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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Hi MHB!

I have decided to extend the deadline by another week so that our members can give this problem another shot and I am looking forward to receive submissions of the solution from the members!(Happy)
 
No one answered to POTW #409.

Here is a suggested solution from other:
View attachment 9675

The function $$S(x)=\sum_{k=1}^{70}\dfrac{k}{x-k}$$ is discontinuous at $x=k$ for $k=1, 2, ..., 70$ but is continuous in the open intervals between these integers. Also, the function goes to $-\infty$ as $x$ approaches $k$ from below and to $+\infty$ as $x$ approaches $k$ from above. Thus the graph crosses the line $y=\dfrac{5}{4}$ in each of the interval $(k, k+1)$ for $k=1, 2, ..., 69$.

For $x>70$, the biggest term in $S(x)$, namely $\dfrac{70}{x-70}$, gets arbitrarily small as $x$ increases, showing that the positive $x$-axis is an asymptote. Therefore from a sketch of the graph of $y=S(x)$, we can see that the set of values of $x$ for which $S(x)\ge \dfrac{5}{4}$ consists of 70 half-open intervals, open on the left and closed on the right, which begin respectively at the integer points $x=1, 2, ..., 70$.

In fact, in the graph of $y=S(x)-\dfrac{5}{4}$, these intervals occurs on the $x$-axis as the intervals between $k$ and the root $x_k$ of the equation $S(x)-\dfrac{5}{4}=0$ which lies between $k$ and $k+1$. The length of the $k$th interval, then, is simply $x_k-k$, and the sum of all 70 intervals is

$(x_1-1)+(x_2-2)+\cdots+(x_{70}-70)=(x_1+x_2+\cdots+x_{70})-(1+2+\cdots+70)$

It remains, then, only to show that the sum of the roots of $S(x)-\dfrac{5}{4}=0$ is $1988+(1+2+\cdots+70)$.

If $S(x)-\dfrac{5}{4}=ax^{70}+bx^{69}+\cdots$, the sum of the roots is simply $-\dfrac{b}{a}$.

$S(x)-\dfrac{5}{4}=\dfrac{1}{x-1}+\dfrac{2}{x-2}+\cdots+\dfrac{70}{x-70}-\dfrac{5}{4}=0$

Clearing the fraction we get

$4(x-2)(x-3)\cdots(x-70)+4\cdot2(x-1)(x-3)\cdots(x-70)+\cdots+4\cdot70(x-1)(x-2)\cdots(x-69)-5(x-1)(x-2)\cdots(x-70)=0$

$-5x^{70}+x^{69}[4\cdot1+4\cdot2+\cdots+4\cdot70-5(-1-2-\cdots-70)]+\cdots=0$

$-5x^{70}+x^{69}[4(1+2+\cdots+70)+5(1+2+\cdots+70)]+\cdots=0$

$-5x^{70}+9(1+2+\cdots+70)x^{69}+\cdots=0$$\begin{align*}-\dfrac{b}{a}&=-\dfrac{9(1+2+\cdots+70)}{-5}\\&=(1+2+\cdots+70)+\dfrac{4(1+2+\cdots+70)}{5}\\&=(1+2+\cdots+70)+\dfrac{4}{5}\dfrac{70}{2}(1+70)\\&=(1+2+\cdots+70)+1988\text{ (Q.E.D.)}\end{align*}$
 

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