請問怎樣用數學歸納法證明這個數列不等式 已知an+1(指第n+1項)=an+(an^2)/(n^2),a1=1/3.求證an>1/2 -1/4n .另外我將不等式放縮成更嚴格的先求證an>1/2 - 1/5n後反而好證了,是什麼道理呢?

請問怎樣用數學歸納法證明這個數列不等式 已知an+1(指第n+1項)=an+(an^2)/(n^2),a1=1/3.求證an>1/2 -1/4n .另外我將不等式放縮成更嚴格的先求證an>1/2 - 1/5n後反而好證了,是什麼道理呢?

證明:
當n=1時,a2=a1+(a1^2)/1^2=1/3+1/18=7/18>1/2-1/4=7/28成立
設n=k時,ak+1=ak+(ak^2)/(k^2)>1/2-1/4k成立
當n=k+1時,ak+2=ak+1+(ak+1^2)/(k+1)^2
>1/2-1/4k+(1/2-1/4k)^2/(k+1)^2
=1/2-1/4[k-(1/2-1/4k)^2/(k+1)^2]
=1/2-1/4[(k^3+2k^2+k-1+k-1/4k^2)/(k+1)^2]
>1/2-1/4[(k^3+3k^2+3k+1)/(k+1)^2]
=1/2-1/4[(k+1)^3/(k+1)^2]
=1/2-1/4(k+1)成立
所以an>1/2 -1/4n