现在取 r=2r=2r=2。若 an+1≤ana_{n+1}\le a_nan+1≤an,则 an+2≤an2+2an+1≤an2+2an+1=an+1a_{n+2}\le\sqrt{a_n^2+2a_{n+1}}\le\sqrt{a_n^2+2a_n+1}=a_n+1an+2≤an2+2an+1≤an2+2an+1=an+1。由于 an+2a_{n+2}an+2 是整数且 an≤an+2<an+1a_n\le a_{n+2}<a_n+1an≤an+2<an+1,必有 an+2=ana_{n+2}=a_nan+2=an。