>>570
(1) log(n)/log(n+1)-1=log(n/(n+1))/log(n+1)→0(n→∞) つまり

(2)　f(x)=sin(1/logx)とおくとf'(x)=-cos(1/logx)/{x(logx)^2}
中間値の定理より
sin(1/log(n))-sin(1/log(n+1))=cos(1/log(n+θ)/{(n+θ)(log(n+θ))^2} (0<θ<1)
なるθが存在する
(1)と同様に
(log(n))^2/(log(n+θ))^2=1
つまり