问题 解答题
已知正项数列{an}满足:
an
-
an-1
=1,(n∈N+,n≥2),且a1=4.
(1)求{an}的通项公式;
(2)求证
1
a1
+
1
a2
+…+
1
an
<1(n∈N+
答案

(1)由题意可知数列{

an
}是等差数列,首项是2,公差为1;

an
=2+(n-1)×1=n+1

∴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
an
<1-
1
2
+
1
2
-
1
3
+…+
1
n
-
1
n+1
=1-
1
n+1
<1

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