Hn=1+1/2+1/3+â¦+1/n-ln(n)>ln(1+1)+ln(1+1/2)+ln(1+1/3)+â¦+ln (1+1/n)-ln(n)
=ln(n+1)-ln(n)=ln(1+1/n)
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Hn-H(n+1)=1+1/2+1/3+â¦+1/n-ln(n)-[1+1/2+1/3+â¦+1/(n+1)-ln(n+1)]
=ln(n+1)-ln(n)-1/(n+1)=ln(1+1/n)-1/(n+1)>ln(1+1/n)-1/n>0
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