设f(n)=1/(n+1)+1/(n+2)+.+1/(n+2^n),则f(k+1)-f(k)=要有详细过程

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设f(n)=1/(n+1)+1/(n+2)+.+1/(n+2^n),则f(k+1)-f(k)=要有详细过程

设f(n)=1/(n+1)+1/(n+2)+.+1/(n+2^n),则f(k+1)-f(k)=要有详细过程
设f(n)=1/(n+1)+1/(n+2)+.+1/(n+2^n),则f(k+1)-f(k)=
要有详细过程

设f(n)=1/(n+1)+1/(n+2)+.+1/(n+2^n),则f(k+1)-f(k)=要有详细过程
f(k)=1/(k+1)+1/(k+2)+...+1/(k+2^k)
f(k+1)=1/(k+2)+1/(k+3)+...+1/(k+2^k)+...+1/[k+1+2^(k+1)]
所以:f(k+1)-f(k)=1/(k+1+2^k)+1/(k+2+2^k)+...+1/[k+1+2^(k+1)]-1/(k+1)