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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)

letSn=p=n+1kn1pSn=p=1kn1pp=1n1p=HknHnwehaveHkn=ln(kn)+γ+o(1n)alsoHn=ln(n)+γ+o(1n)(n+)HknHn=ln(k)+o(1n)limn+Sn=ln(k)

Commented by M±th+et£s last updated on 19/Feb/20

thank you sir

thankyousir

Commented by mathmax by abdo last updated on 19/Feb/20

you are welcome.

youarewelcome.

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