(1/n^2)納k=1,n]k*ln(n+k)-(1/2)ln(n)
=(1/n^2)[納k=1,n]{k*ln(n+k)-k*ln(n)}+(1/2)n(n+1)ln(n)] - (1/2)ln(n)
={(1/n)納k=1,n](k/n)*ln(1+k/n)}+(1/2)(1/n)ln(n)
→ ∫[0,1] ln(x+1)dx = [(x+1)ln(x+1)][0,1]-∫[0,1]dx = 2ln(2) -1