Problem 2

Let

f(x)

and

g(x)

be given by

f(x)=\dfrac1x+\dfrac1{x-2}+\dfrac1{x-4}+\cdots+\dfrac1{x-2018}

and

g(x)=\dfrac1{x-1}+\dfrac1{x-3}+\dfrac1{x-5}+\cdots+\dfrac1{x-2017}

. Prove that

|f(x)-g(x)|>2

for any non-integer real number

x

satisfying

0<x<2018

Step 3 of 6: Shifting by two does not decrease d

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d(x)=g(x)-f(x),\qquad d(x+2)-d(x)=\Big(\frac1{x+1}-\frac1{x+2}\Big)-\Big(\frac1{x-2017}-\frac1{x-2018}\Big)\ge 0

Detailed analysis

Write $d=g-f$ . Reindexing the sums shows $d(x+2)-d(x)$ telescopes to exactly the displayed difference of two brackets. For $x$ in the relevant range the first bracket is positive since $x+1<x+2$ , and the second bracket is non-positive since both $x-2017$ and $x-2018$ are large negative numbers with $x-2018$ more negative; hence $d(x+2)\ge d(x)$ . So on the family $2n-1<x<2n$ , the value of $d$ is smallest when $n=1$ , i.e. it suffices to prove $d(x)>2$ for $1<x<2$ .