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Let-f-x-and-g-x-be-given-by-f-x-1-x-1-x-2-1-x-4-1-x-2018-and-g-x-1-x-1-1-x-3-1-x-5-1-x-2017-Prove-that-f-x-g-x-gt-2-for-any-non-in

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Question Number 201441 by dimentri last updated on 06/Dec/23
Let f(x) and g(x) be given by f(x)= (1/x) +(1/(x−2)) +(1/(x−4)) + ... +(1/(x−2018)) and g(x)=(1/(x−1)) +(1/(x−3)) +(1/(x−5)) +...+ (1/(x−2017)). Prove that ∣ f(x)−g(x)∣ >2 for any non−integer real number x satisfying 0<x<2018.
$${Let}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{be}\:{given}\:{by}\: \\ $$$$\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}−\mathrm{2}}\:+\frac{\mathrm{1}}{{x}−\mathrm{4}}\:+\:…\:+\frac{\mathrm{1}}{{x}−\mathrm{2018}} \\ $$$$\:{and}\: \\ $$$$\:\:{g}\left({x}\right)=\frac{\mathrm{1}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{{x}−\mathrm{3}}\:+\frac{\mathrm{1}}{{x}−\mathrm{5}}\:+…+\:\frac{\mathrm{1}}{{x}−\mathrm{2017}}. \\ $$$$\:\:{Prove}\:{that}\:\:\mid\:{f}\left({x}\right)−{g}\left({x}\right)\mid\:>\mathrm{2} \\ $$$$\:\:{for}\:{any}\:{non}−{integer}\:{real}\:{number} \\ $$$$\:\:{x}\:{satisfying}\:\mathrm{0}<{x}<\mathrm{2018}.\: \\ $$
Answered by Rasheed.Sindhi last updated on 06/Dec/23
∣ f(x)−g(x) ∣= ∣ (((−1)^(1+1) )/(x+1−1))+(((−1)^(2+1) )/(x+1−2))+(((−1)^(3+1) )/(x+1−3))+...+(((−1)^(2018+1) )/(x+1−2018))+(((−1)^(2019+1) )/(x+1−2019)) ∣ =∣ Σ_(n=1) ^(2019) ((((−1)^(n+1) )/(x−n+1))) ∣ ....
$$\mid\:{f}\left({x}\right)−{g}\left({x}\right)\:\mid= \\ $$$$\mid\:\frac{\left(−\mathrm{1}\right)^{\mathrm{1}+\mathrm{1}} }{{x}+\mathrm{1}−\mathrm{1}}+\frac{\left(−\mathrm{1}\right)^{\mathrm{2}+\mathrm{1}} }{{x}+\mathrm{1}−\mathrm{2}}+\frac{\left(−\mathrm{1}\right)^{\mathrm{3}+\mathrm{1}} }{{x}+\mathrm{1}−\mathrm{3}}+…+\frac{\left(−\mathrm{1}\right)^{\mathrm{2018}+\mathrm{1}} }{{x}+\mathrm{1}−\mathrm{2018}}+\frac{\left(−\mathrm{1}\right)^{\mathrm{2019}+\mathrm{1}} }{{x}+\mathrm{1}−\mathrm{2019}}\:\mid \\ $$$$=\mid\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{2019}} {\Sigma}}\:\:\left(\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{x}−{n}+\mathrm{1}}\right)\:\mid \\ $$$$…. \\ $$

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