问题
解答题
已知正项数列{an}满足:
(1)求{an}的通项公式; (2)求证
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答案
(1)由题意可知数列{
}是等差数列,首项是2,公差为1;an
∴
=2+(n-1)×1=n+1an
∴an=(n+1)2
(2)证明:
=1 ak
<1 (k+1)2
=1 k(k+1)
-1 k 1 k+1
∴
+1 a1
+…+1 a2
<1-1 an
+1 2
-1 2
+…+1 3
-1 n
=1-1 n+1
<11 n+1