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Question Number 30194 by abdo imad last updated on 17/Feb/18
$${study}\:{the}\:{convergence}\:{of}\:{u}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{{C}_{{n}} ^{{k}} }\:\:{with} \\ $$$${C}_{{n}} ^{{k}} \:\:=\frac{{n}!}{{k}!\left({n}−{k}\right)!}\:. \\ $$
Commented by prof Abdo imad last updated on 22/Feb/18
$${let}\:{put}\:\:{x}_{{n}} =\:\frac{\mathrm{1}}{{C}_{{n}} ^{{k}} }\:\:\:\:\forall\:{n}\in\:{N}\:\:{x}_{{n}} >\mathrm{0}\:\:{we}\:{have} \\ $$$$ \\ $$$$\frac{{x}_{{n}+\mathrm{1}} }{{x}_{{n}} }\:=\:\:\frac{{C}_{{n}} ^{{k}} }{{C}_{{n}+\mathrm{1}} ^{{k}} }\:=\:\frac{\frac{{n}!}{{k}!\left({n}−{k}\right)!}}{\frac{\left({n}+\mathrm{1}\right)!}{{k}!\left({n}+\mathrm{1}−{k}\right)!}}=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\frac{\left({n}+\mathrm{1}−{k}\right)!}{\left({n}−{k}\right)!} \\ $$$$=\:\frac{{n}+\mathrm{1}−{k}}{{n}+\mathrm{1}}=\:\mathrm{1}−\:\frac{{k}}{{n}+\mathrm{1}}\:<\mathrm{1}\:\:{from}\:{that}\:\:{u}_{{n}} \:{isconvergent}. \\ $$