1.當(dāng)n=2時(shí),1+根號(hào)2>根號(hào)2,顯然成立.
假設(shè)n=k時(shí)成立,即1+1/根號(hào)2+1/根號(hào)3+…+1/根號(hào)k>根號(hào)k
當(dāng)n=k+1時(shí),左=1+1/根號(hào)2+1/根號(hào)3+…
+1/根號(hào)k+1/根號(hào)(k+1)
>根號(hào)k+1/根號(hào)(k+1)
根號(hào)k+1/根號(hào)(k+1)-根號(hào)(k+1)
={[根號(hào)k*根號(hào)(k+1)]-k}/根號(hào)(k+1)
由于k>2,根號(hào)k*根號(hào)(k+1)]-k>0
根號(hào)k+1/根號(hào)(k+1)-根號(hào)(k+1)>0
根號(hào)k+1/根號(hào)(k+1)>根號(hào)(k+1)
即當(dāng)n=k+1時(shí),左=1+1/根號(hào)2+1/根號(hào)3+…
+1/根號(hào)k+1/根號(hào)(k+1)
>根號(hào)k+1/根號(hào)(k+1)>根號(hào)(k+1)=右邊
從而命題成立