Question Number 215979 by alcohol last updated on 23/Jan/25

$${u}_{{n}} \:=\:\underset{{k}={n}+\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:{and}\:{v}_{{n}} \:=\:\underset{{k}={n}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$$$\bullet\:{show}\:{that}\:{u}_{{n}} \:{and}\:{v}_{{n}} \:{are}\:{adjacent} \\ $$$${use}\:{ln}\left({x}+\mathrm{1}\right)\:\leqslant\:{x}\:{and}\:{x}\leqslant−{ln}\left(\mathrm{1}−{x}\right)\:{and} \\ $$$$\bullet\:{show}\:{that}\:{u}_{{n}} \:\leqslant\:\underset{{k}={n}+\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\left({ln}\left({k}\right)−{ln}\left({k}−\mathrm{1}\right)\right) \\ $$$${hence}\:{deduce}\:{that}\:{u}_{{n}} \:\leqslant\:{ln}\mathrm{2} \\ $$$$\bullet\:{show}\:{that}\:{v}_{{n}} \:\geqslant\:\underset{{k}={n}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\left({ln}\left({k}+\mathrm{1}\right)−{ln}\left({k}\right)\right) \\ $$$${hence}\:{deduce}\:{that}\:{v}_{{n}} \geqslant{ln}\mathrm{2} \\ $$