Question Number 212627 by MrGaster last updated on 19/Oct/24
$$ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\sqrt{\mathrm{1}\centerdot\mathrm{2}}}{{n}^{\mathrm{2}} +\mathrm{1}}+\frac{\sqrt{\mathrm{2}\centerdot\mathrm{3}}}{{n}^{\mathrm{2}} +\mathrm{2}}+\ldots+\frac{\sqrt{{n}\left({n}+\mathrm{1}\right)}}{{n}^{\mathrm{2}} +{n}}\right) \\ $$
Answered by mehdee7396 last updated on 19/Oct/24
$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}}+\frac{\mathrm{2}}{{n}^{\mathrm{2}} +{n}}+\frac{\mathrm{3}}{{n}^{\mathrm{2}} +{n}}+…+\frac{{n}}{{n}^{\mathrm{2}} +{n}}<{S}_{{n}} \:\: \\ $$$$\Rightarrow\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}\left({n}^{\mathrm{2}} +{n}\right)}<{S}_{{n}} \\ $$$${S}_{{n}} <\frac{\mathrm{2}}{{n}^{\mathrm{2}} }+\frac{\mathrm{3}}{{n}^{\mathrm{2}} }+\frac{\mathrm{4}}{{n}^{\mathrm{2}} }+…+\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\Rightarrow{S}_{{n}} <\frac{{n}\left({n}+\mathrm{3}\right)}{\mathrm{2}{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}\left({n}^{\mathrm{2}} +{n}\right)}<{S}_{{n}} <\frac{{n}\left({n}+\mathrm{3}\right)}{\mathrm{2}{n}^{\mathrm{2}} } \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} {S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:\:\checkmark \\ $$$$ \\ $$