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Question Number 141409 by Willson last updated on 18/May/21

Prove that ∀n∈N ∫^( n+1) _n lnt dt ≤ ln(n+(1/2))

$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\forall\mathrm{n}\in\mathbb{N}\:\:\:\:\underset{\mathrm{n}} {\int}^{\:\mathrm{n}+\mathrm{1}} \mathrm{lnt}\:\mathrm{dt}\:\leqslant\:\mathrm{ln}\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$

Answered by TheSupreme last updated on 19/May/21

∫ln(t)=xln(t)−x ∫_n ^(n+1) ...=(n+1)ln(n+1)−n−1−nln(n)+n =nln(n+1)+ln(n+1)−nln(n)−1 =ln(1+(1/n))^n +ln(n+1)−1 for n>1 ln(t) is monotone and positive so ∫ is monotone max(∫)=lim_x ln(1+(1/n))^n +ln(n+1)−1= =1+ln(n+1)−1=ln(n+1) ∫_n ^(n+1) ln(t)dt≤ln(n+1)

$$\int{ln}\left({t}\right)={xln}\left({t}\right)−{x} \\ $$$$\int_{{n}} ^{{n}+\mathrm{1}} ...=\left({n}+\mathrm{1}\right){ln}\left({n}+\mathrm{1}\right)−{n}−\mathrm{1}−{nln}\left({n}\right)+{n} \\ $$$$={nln}\left({n}+\mathrm{1}\right)+{ln}\left({n}+\mathrm{1}\right)−{nln}\left({n}\right)−\mathrm{1} \\ $$$$={ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} +{ln}\left({n}+\mathrm{1}\right)−\mathrm{1} \\ $$$${for}\:{n}>\mathrm{1}\:{ln}\left({t}\right)\:{is}\:{monotone}\:{and}\:{positive}\:{so}\:\int\:{is}\:{monotone} \\ $$$${max}\left(\int\right)={li}\underset{{x}} {{m}}\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} +{ln}\left({n}+\mathrm{1}\right)−\mathrm{1}= \\ $$$$=\mathrm{1}+{ln}\left({n}+\mathrm{1}\right)−\mathrm{1}={ln}\left({n}+\mathrm{1}\right) \\ $$$$\int_{{n}} ^{{n}+\mathrm{1}} {ln}\left({t}\right){dt}\leqslant{ln}\left({n}+\mathrm{1}\right) \\ $$

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