Question Number 205772 by mr W last updated on 30/Mar/24
Answered by MM42 last updated on 30/Mar/24
$${S}_{{n}} =\left(\sqrt{\mathrm{1}×\mathrm{2}}−\mathrm{1}\right) \\ $$$$+\left(\sqrt{\mathrm{2}×\mathrm{3}}−\sqrt{\mathrm{1}×\mathrm{2}}−\mathrm{1}\right) \\ $$$$+\left(\sqrt{\mathrm{3}×\mathrm{4}}−\sqrt{\mathrm{2}×\mathrm{3}}−\mathrm{1}\right) \\ $$$$\vdots \\ $$$$+\left(\sqrt{{n}\left({n}+\mathrm{1}\right)}−\sqrt{\left({n}−\mathrm{1}\right){n}}−\mathrm{1}\right) \\ $$$$+\left(\sqrt{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}−\sqrt{{n}\left({n}+\mathrm{1}\right)}−\mathrm{1}\right) \\ $$$$\Rightarrow{S}_{{n}} =\sqrt{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}−{n}−\mathrm{1} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} \:{S}_{{n}} \:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\checkmark \\ $$$$ \\ $$
Answered by A5T last updated on 30/Mar/24
$${T}_{\mathrm{0}} =\sqrt{\mathrm{2}}−\mathrm{0}−\mathrm{1} \\ $$$${T}_{{n}−\mathrm{1}} =\sqrt{{n}^{\mathrm{2}} +{n}}−\sqrt{{n}^{\mathrm{2}} −{n}}−\mathrm{1} \\ $$$${T}_{{n}} =\sqrt{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}−\sqrt{{n}^{\mathrm{2}} +{n}}−\mathrm{1} \\ $$$${S}_{{n}} =−\left({n}+\mathrm{1}\right)+\sqrt{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}} \\ $$$${n}+\mathrm{1}+{n}+\mathrm{2}>\mathrm{2}\sqrt{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\Rightarrow\sqrt{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}<\frac{\mathrm{2}{n}+\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{S}_{{n}} =\sqrt{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}−{n}−\mathrm{1}<{n}+\frac{\mathrm{3}}{\mathrm{2}}−{n}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Question}\Rightarrow\underset{{n}\rightarrow\infty} {{lim}S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$