Question

ln¥1

[¥]求证：ln{(1+1·2)(1+2·3)…[1+n(n+1)]}>nlnn (n为正整数)

Answer 1

∵1+i+i²=(1/2+i)²+3/4>(1/2+i)²,∴ln(1+i+i²)>2ln(1/2+i)
∴ln{(1+1·2)(1+2·3)…[1+n(n+1)]}=2Σln(1/2+i),(i=1,.n)
∴2*ln{(1+1·2)(1+2·3)…[1+n(n+1)]}=2Σ(ln(1/2+i)+ln(1/2+n-i+1))=2Σ(ln(1/2+i)(1/2+n-i+1)),(i=1,.n,倒叙相加)
∵(1/2+i)(1/2+n-i+1)=(n+1-i)*i+n/2+3/4,而(n+1-i)*i≥n(在i=1或n时取最小值)
∴(1/2+i)(1/2+n-i+1)>n
∴2*ln{(1+1·2)(1+2·3)…[1+n(n+1)]}=2Σ(ln(1/2+i)(1/2+n-i+1))>2Σlnn=2nlnn
∴ln{(1+1·2)(1+2·3)…[1+n(n+1)]}>nlnn