Question Number 141409 by Willson last updated on 18/May/21
$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\forall\mathrm{n}\in\mathbb{N}\:\:\:\:\underset{\mathrm{n}} {\int}^{\:\mathrm{n}+\mathrm{1}} \mathrm{lnt}\:\mathrm{dt}\:\leqslant\:\mathrm{ln}\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$
Answered by TheSupreme last updated on 19/May/21
$$\int{ln}\left({t}\right)={xln}\left({t}\right)−{x} \\ $$$$\int_{{n}} ^{{n}+\mathrm{1}} …=\left({n}+\mathrm{1}\right){ln}\left({n}+\mathrm{1}\right)−{n}−\mathrm{1}−{nln}\left({n}\right)+{n} \\ $$$$={nln}\left({n}+\mathrm{1}\right)+{ln}\left({n}+\mathrm{1}\right)−{nln}\left({n}\right)−\mathrm{1} \\ $$$$={ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} +{ln}\left({n}+\mathrm{1}\right)−\mathrm{1} \\ $$$${for}\:{n}>\mathrm{1}\:{ln}\left({t}\right)\:{is}\:{monotone}\:{and}\:{positive}\:{so}\:\int\:{is}\:{monotone} \\ $$$${max}\left(\int\right)={li}\underset{{x}} {{m}}\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} +{ln}\left({n}+\mathrm{1}\right)−\mathrm{1}= \\ $$$$=\mathrm{1}+{ln}\left({n}+\mathrm{1}\right)−\mathrm{1}={ln}\left({n}+\mathrm{1}\right) \\ $$$$\int_{{n}} ^{{n}+\mathrm{1}} {ln}\left({t}\right){dt}\leqslant{ln}\left({n}+\mathrm{1}\right) \\ $$