令f(x)=a^(1/x),则f'(x)=-(1/x²)(a^(1/x))·lna,由中值定理知
存在ξ∈(n,n+1),使得f'(ξ)=f(n+1)-f(n)
即a^(1/(n+1))-a^(1/n)=-(1/ξ²)(a^(1/ξ))·lna
=>[a^(1/(n+1))-a^(1/n)]/lna=(1/ξ²)a^(1/ξ)
∵n
令f(x)=a^(1/x),则f'(x)=-(1/x²)(a^(1/x))·lna,由中值定理知
存在ξ∈(n,n+1),使得f'(ξ)=f(n+1)-f(n)
即a^(1/(n+1))-a^(1/n)=-(1/ξ²)(a^(1/ξ))·lna
=>[a^(1/(n+1))-a^(1/n)]/lna=(1/ξ²)a^(1/ξ)
∵n