为什么N/(N^2 + 1)^(1/2) > 1/(N^2 + 1)^(1/2) + 1/(N^2 + 2)^(1/2) + ... + 1/(N^2 + N)^(1/2)

因为(N^2 + 1)^(1/2)<(N^2 + 2)^(1/2)<(N^2 + 3)^(1/2)<(N^2 + 4)^(1/2)<……<(N^2 + N)^(1/2)
所以
1/(N^2 + 1)^(1/2) > 1/(N^2 + 2)^(1/2) > ... > 1/(N^2 + N)^(1/2)
而N/(N^2 + 1)^(1/2)=1/(N^2 + 1)^(1/2)+1/(N^2 + 1)^(1/2)+1/(N^2 + 1)^(1/2)+1/(N^2 + 1)^(1/2)+……1/(N^2 + 1)^(1/2)
所以N/(N^2 + 1)^(1/2) > 1/(N^2 + 1)^(1/2) + 1/(N^2 + 2)^(1/2) + ... + 1/(N^2 + N)^(1/2)
与上面同理(N^2 + 1)^(1/2)<(N^2 + 2)^(1/2)<(N^2 + 3)^(1/2)<(N^2 + 4)^(1/2)<……<(N^2 + N)^(1/2)<1/(N^2 + N)^(1/2)
所以 1/(N^2 + 1)^(1/2) + 1/(N^2 + 2)^(1/2) + ... + 1/(N^2 + N)^(1/2) > 1/(N^2 + N)^(1/2)+1/(N^2 + N)^(1/2)+1/(N^2 + N)^(1/2)…+1/(N^2 + N)^(1/2)=N/(N^2 + N)^(1/2)

综上所述N/(N^2 + 1)^(1/2) > 1/(N^2 + 1)^(1/2) + 1/(N^2 + 2)^(1/2) + ... + 1/(N^2 + N)^(1/2) > N/(N^2 + N)^(1/2)

因为N/(N^2 + 1)^(1/2) 是N个1/(N^2 + 1)^(1/2)，和第二个式子相比只有第一项相等其余都是大于第二项的各个分式的（因为分母小），同理可得后面的也是一样的