令f(x)=ln(1+x)-[x/(1+x)],x∈(0,1]
f'(x)=[1/(1+x)]-[1/(1+x)²]=x/(1+x)²>0,故f(x)在(0,1]遞增,∴函數(shù)f(x)>f(0)=0
則ln(1+x)>x/(1+x),x∈(0,1]
令x=1/n,則ln[1+(1/n)]>1/(n+1),n≥1,且n∈N*
即ln[(n+1)/n]>1/(n+1),∴l(xiāng)n(n+1)-lnn>1/(n+1)
ln2-ln1+ln3-ln2+...+ln(n+1)-lnn>1/2 + 1/3 +...+1/(n+1)
∴l(xiāng)n(n+1)-ln1>1/2 + 1/3 +.+1/(n+1),即得證.