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Question Number 26111 by abdo imad last updated on 19/Dec/17
$${find}\:\:{the}\:{radius}\:{of}\:{convergence}\:{for}\:{the}\:{serie}\:\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:{H}_{{n}} \:{x}^{{n}} \\ $$$${H}_{{n}} \:\:=\:\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}}\:. \\ $$
Commented by prakash jain last updated on 21/Dec/17
$$\mid{x}\mid<\mathrm{1} \\ $$
Commented by abdo imad last updated on 22/Dec/17
$${we}\:{put}\:\:{a}_{{n}^{} } ={H}_{{n}} \:\:\:>\mathrm{0}\:\:{lim}_{{n}−>\propto} \frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }\:={lim}_{{n}−>\propto} \frac{{H}_{{n}+\mathrm{1}} }{{H}_{{n}} } \\ $$$$={lim}_{{n}−>\propto^{} } \:\frac{{H}_{{n}} \:+\:\frac{\mathrm{1}}{{n}+\mathrm{1}}}{{H}_{{n}} }=={lim}_{{n}−>\propto} \left(\mathrm{1}+\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){H}_{{n}} }\right)=\mathrm{1} \\ $$$${because}\:{H}_{{n}} \:\:\sim_{{n}−>\propto} \:{ln}\left({n}\right)−>\propto \\ $$$${we}\:{have}\:\:\frac{\mathrm{1}}{{R}}={lim}\:_{{n}−>\propto} \frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\mathrm{1}\:\:\Rightarrow\:\:\:{R}=\mathrm{1} \\ $$