Question Number 127430 by mathocean1 last updated on 29/Dec/20
$${The}\:{objective}\:{of}\:{this}\:{exercise} \\ $$$${is}\:{to}\:{calculate}\:\underset{{n}\rightarrow+\infty} {{lim}}\frac{\mathrm{1}}{\sqrt{{n}}}\:×\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\sqrt{{n}+{k}}}. \\ $$$${Given}\:{S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\sqrt{\mathrm{k}}};\:{U}_{{n}} =\mathrm{2}\sqrt{\mathrm{2}}−{S}_{{n}} \\ $$$${and}\:{V}_{{n}} =\mathrm{2}\sqrt{{n}+\mathrm{1}}−{Sn}. \\ $$$$\mathrm{1}.\:{show}\:\:{that}\underset{{n}\rightarrow+\infty} {{lim}Sn}=+\infty. \\ $$$$\mathrm{2}.\:{show}\:{that}\:{V}_{{n}} \:{and}\:{U}_{{n}} \:{are}\: \\ $$$${adjacent}\:{then}\:{deduct}\:{that}\:\:{their} \\ $$$${common}\:{limit}\:{is}\:{L}\geqslant\mathrm{1}. \\ $$$$\mathrm{3}.\:{Calculate}\:\underset{{n}\rightarrow+\infty} {{lim}}\left(\frac{\mathrm{S}_{{n}} }{{n}}\right)\:{and} \\ $$$$\underset{{n}\rightarrow+\infty} {{lim}}\left(\frac{{S}_{{n}} }{\sqrt{{n}}}\right). \\ $$$$\mathrm{4}.\:{Deduct}\:{from}\:{last}\:{questions} \\ $$$$\:\underset{{n}\rightarrow+\infty} {{lim}}\frac{\mathrm{1}}{\sqrt{{n}}}\:×\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\sqrt{{n}+{k}}}. \\ $$
Commented by mathocean1 last updated on 29/Dec/20
$${sorry} \\ $$
Commented by talminator2856791 last updated on 29/Dec/20
$${hahahaaa} \\ $$$$ \\ $$