Menu Close

u-n-k-n-1-2n-1-k-and-v-n-k-n-2n-1-1-k-show-that-u-n-and-v-n-are-adjacent-use-ln-x-1-x-and-x-ln-1-x-and-show-that-u-n-k-n-1-2n-ln-k-ln-k-1-hence-deduce-




Question Number 215979 by alcohol last updated on 23/Jan/25
u_n  = Σ_(k=n+1) ^(2n) (1/k) and v_n  = Σ_(k=n) ^(2n−1) (1/k)  • show that u_n  and v_n  are adjacent  use ln(x+1) ≤ x and x≤−ln(1−x) and  • show that u_n  ≤ Σ_(k=n+1) ^(2n) (ln(k)−ln(k−1))  hence deduce that u_n  ≤ ln2  • show that v_n  ≥ Σ_(k=n) ^(2n−1) (ln(k+1)−ln(k))  hence deduce that v_n ≥ln2
$${u}_{{n}} \:=\:\underset{{k}={n}+\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:{and}\:{v}_{{n}} \:=\:\underset{{k}={n}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$$$\bullet\:{show}\:{that}\:{u}_{{n}} \:{and}\:{v}_{{n}} \:{are}\:{adjacent} \\ $$$${use}\:{ln}\left({x}+\mathrm{1}\right)\:\leqslant\:{x}\:{and}\:{x}\leqslant−{ln}\left(\mathrm{1}−{x}\right)\:{and} \\ $$$$\bullet\:{show}\:{that}\:{u}_{{n}} \:\leqslant\:\underset{{k}={n}+\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\left({ln}\left({k}\right)−{ln}\left({k}−\mathrm{1}\right)\right) \\ $$$${hence}\:{deduce}\:{that}\:{u}_{{n}} \:\leqslant\:{ln}\mathrm{2} \\ $$$$\bullet\:{show}\:{that}\:{v}_{{n}} \:\geqslant\:\underset{{k}={n}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\left({ln}\left({k}+\mathrm{1}\right)−{ln}\left({k}\right)\right) \\ $$$${hence}\:{deduce}\:{that}\:{v}_{{n}} \geqslant{ln}\mathrm{2} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *