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Question-82225




Question Number 82225 by M±th+et£s last updated on 19/Feb/20
Commented by mathmax by abdo last updated on 19/Feb/20
let S_n =Σ_(p=n+1) ^(kn)  (1/p) ⇒S_n =Σ_(p=1) ^(kn)  (1/p)−Σ_(p=1) ^n  (1/p) =H_(kn) −H_n   we have H_(kn) =ln(kn)+γ +o((1/n)) also H_n =ln(n)+γ +o((1/n))(n→+∞)  H_(kn) −H_n =ln(k)+o((1/n)) ⇒lim_(n→+∞)  S_n =ln(k)
$${let}\:{S}_{{n}} =\sum_{{p}={n}+\mathrm{1}} ^{{kn}} \:\frac{\mathrm{1}}{{p}}\:\Rightarrow{S}_{{n}} =\sum_{{p}=\mathrm{1}} ^{{kn}} \:\frac{\mathrm{1}}{{p}}−\sum_{{p}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{p}}\:={H}_{{kn}} −{H}_{{n}} \\ $$$${we}\:{have}\:{H}_{{kn}} ={ln}\left({kn}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:{also}\:{H}_{{n}} ={ln}\left({n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\left({n}\rightarrow+\infty\right) \\ $$$${H}_{{kn}} −{H}_{{n}} ={ln}\left({k}\right)+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} ={ln}\left({k}\right) \\ $$
Commented by M±th+et£s last updated on 19/Feb/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 19/Feb/20
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$

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