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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
letSn=∑p=n+1kn1p⇒Sn=∑p=1kn1p−∑p=1n1p=Hkn−HnwehaveHkn=ln(kn)+γ+o(1n)alsoHn=ln(n)+γ+o(1n)(n→+∞)Hkn−Hn=ln(k)+o(1n)⇒limn→+∞Sn=ln(k)
Commented by M±th+et£s last updated on 19/Feb/20
thankyousir
youarewelcome.
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