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Question Number 74795 by mathmax by abdo last updated on 30/Nov/19
$${study}\:{the}\:{convergence}\:{of}\:\Sigma\:\frac{\mathrm{1}}{{nH}_{{n}} } \\ $$$${with}\:{H}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}} \\ $$
Commented by mathmax by abdo last updated on 01/Dec/19
$${we}\:{have}\:\:{H}_{{n}} \sim\:{ln}\left({n}\right)\:\:\:\left({n}\rightarrow+\infty\right)\:\Rightarrow\frac{\mathrm{1}}{{n}\:{H}_{{n}} }\:\sim\:\frac{\mathrm{1}}{{nln}\left({n}\right)} \\ $$$${let}\:{U}_{{n}} =\frac{\mathrm{1}}{{nln}\left({n}\right)}\:\:{we}\:{have}\:\:{U}_{{n}} \:{decrease}\:{to}\:\mathrm{0}\:\:{so}\:\int_{\mathrm{2}} ^{+\infty} \:\frac{{dt}}{{tln}\left({t}\right)}\:{and} \\ $$$$\Sigma\:{U}_{{n}} \:{have}\:{same}\:{nature}\:\:\:{changement}\:{ln}\left({t}\right)={u}\:{give} \\ $$$$\int_{\mathrm{2}} ^{+\infty} \:\frac{{dt}}{{tln}\left({t}\right)}\:=\int_{{ln}\left(\mathrm{2}\right)} ^{+\infty} \:\:\:\frac{{e}^{{u}} \:{du}}{{e}^{{u}} \:{u}}\:=\int_{{ln}\left(\mathrm{2}\right)} ^{+\infty} \:\frac{{du}}{{u}}\:=+\infty\:\Rightarrow\Sigma\:{U}_{{n}} {diverges}\:\Rightarrow \\ $$$$\Sigma\:\frac{\mathrm{1}}{{nH}_{{n}} }\:\:{diverges}. \\ $$
Answered by mind is power last updated on 01/Dec/19
$$\Sigma\frac{\mathrm{1}}{\mathrm{nH}_{\mathrm{n}} } \\ $$$$\mathrm{H}_{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\leqslant\int_{\mathrm{k}} ^{\mathrm{k}+\mathrm{1}} \frac{\mathrm{1}}{\mathrm{t}}\leqslant\frac{\mathrm{1}}{\mathrm{k}} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\leqslant\mathrm{ln}\left(\mathrm{n}+\mathrm{1}\right)\leqslant\mathrm{H}_{{n}} \\ $$$$\Rightarrow\mathrm{H}_{\mathrm{n}} \geqslant\mathrm{ln}\left(\mathrm{n}+\mathrm{1}\right)\Rightarrow\frac{\mathrm{1}}{\mathrm{nH}_{\mathrm{n}} }\leqslant\frac{\mathrm{1}}{\mathrm{nln}\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\:\:\:\mathrm{H}_{\mathrm{n}+\mathrm{1}} −\mathrm{1}\leqslant\mathrm{ln}\left(\mathrm{n}+\mathrm{1}\right) \\ $$$$\mathrm{H}_{\mathrm{n}} \leqslant\mathrm{ln}\left(\mathrm{n}\right)+\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\mathrm{nH}_{\mathrm{n}} }\geqslant\frac{\mathrm{1}}{\mathrm{nln}\left(\mathrm{n}\right)+\mathrm{n}} \\ $$$$\Rightarrow\Sigma\frac{\mathrm{1}}{\mathrm{nH}_{\mathrm{n}} }\geqslant\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{ln}\left(\mathrm{n}\right)+\mathrm{1}\right)}\rightarrow+\infty \\ $$$$ \\ $$