设f(n)=1+1/2+1/3...+1/n,求f(2^(k+1))-f(2^k)=?

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设f(n)=1+1/2+1/3...+1/n,求f(2^(k+1))-f(2^k)=?

设f(n)=1+1/2+1/3...+1/n,求f(2^(k+1))-f(2^k)=?
设f(n)=1+1/2+1/3...+1/n,求f(2^(k+1))-f(2^k)=?

设f(n)=1+1/2+1/3...+1/n,求f(2^(k+1))-f(2^k)=?
f(2^(k+1))-f(2^k)
=1+1/2+/1/3+...+1/2^k+1/(2^k+1)+1/(2^k+2)+...+1/2^(k+1)-1-1/2-1/3-...-1/2^k
=1/(2^k+1)+1/(2^k+2)+...+1/2^(k+1)