Previous in Relation and Functions Next in Relation and Functions
Question Number 144218 by Mathspace last updated on 23/Jun/21
$${U}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{k}+\mathrm{1}}} \\ $$$${find}\:{a}\:{eqivalent}\:{of}\:{U}_{{n}} \left({n}\rightarrow\infty\right) \\ $$
Answered by ArielVyny last updated on 24/Jun/21
$$\mathrm{0}\leqslant{k}\leqslant{n}\rightarrow\mathrm{0}\leqslant\mathrm{2}{k}\leqslant\mathrm{2}{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\rightarrow\mathrm{1}\leqslant\mathrm{2}{k}+\mathrm{1}\leqslant\mathrm{2}{n}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{n}+\mathrm{1}}}\leqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{k}+\mathrm{1}}}\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\rightarrow\frac{{n}+\mathrm{1}}{\:\sqrt{\mathrm{2}{n}+\mathrm{1}}}\leqslant{U}_{{n}} \leqslant{n}+\mathrm{1} \\ $$