(1)求{an}的通项公式;
(2)求证:
+
+
+…+
<
.
(1)求{an}的通项公式;
(2)求证:+++…+<.
(1)解:记Sn=a1+a2+…+an,由题意得n≥2时,an=Sn－1.
由
两式相减得an+1－an=an(n≥2),
∴n≥2时,an+1=2an,即a2,a3,a4,…是以2为公比的等比数列.
∵a1=5,a2=5,∴n≥2时,an=5·2n－2,
即an=
(2)证明:+++…+
=
+
+
+…+
=
+
［1－(
)n－1］<
+
=
.