如题所述
求过程
根据之前的假设,当n=k时,1+1/2+1/3+...+1/2k-1<k成立,所以有1+1/2+1/3+...+1/2k-1+1/2k+1/(2k+1)<k+1/2k+1/(2k+1),将k+1/2k+1/(2k+1)通分,得到(4k³+2k²+4k+1)/(4k²+2k),(4k³+2k²+4k+1)/(4k²+2k)<(4k³+6k²+2k)/(4k²+2k)=k+1(→可以证明)∴1+1/2+1/3+...+1/2k-1+1/2k+1/(2k+1)<k+1,得证。