n已知数列{an}满足a1=1/2,an=an-1/an-1+2(n>=2,n属于N).(1)求数列{an}的通项an,

（1）由a = a /(a +2),得
n=1时,a1=1/2
n=2时,a2 = (1/2) / (1/2 + 2) = 1/5 = (2²+1)^(-1)
n=3时,a3 = (1/5) / (1/5 + 2) = 1/11 = (2³+2+1)^(-1)
猜想,当n=k时,a = [ 2^k + 2^(k-2) + 2^(k-3)+……+ 2²+2+ 1 ]^(-1)
= [ 2^k + 2^(k-1) +2^(k-2) + 2^(k-3)+……+ 2²+2+ 1 - 2^(k-1) ]^(-1)
= [ 2^(k+1) - 1 - 2^(k-1) ]^(-1)
= [3*2^(k-1) - 1]^(-1)
满足题意a = a /(a +2).其中,k∈N,且k≥2
当n=k+1时,a = 1 ÷ [ 3*2^k - 1 ]
= （1*a）÷【[ 3*2^k - 1 ] * a】 ←（分子分母同时乘以a）
= a ÷ 【[ 2*3*2^(k-1) - 2 + 1 ] /[3*2^(k-1) - 1]】
= a ÷ 【 2 + 1 /[3*2^(k-1) - 1]】
= a /(2+a)
∴当n=k+1时,原猜想也成立.
且当n=1时,a1 = 1/(3*2^0 - 1) = 1/2 ,原猜想也成立
∴对于任意n∈N,恒有 a = 1/ [ 3*2^(n-1) - 1]
（2）Cn = an/(1+2an)
把an的通项代入上式,化简得
Cn = 1 /[3*2^(n-1) + 1]
令Dn = 1 /[2^(n-3)] = 4*(1/2)^(n-1)
则 1/Cn - 7/Dn = [ 3*2^(n-1) + 1] - [ 7* 2^(n-3) ] = 5*2^(n-3) + 1 ＞ 0
∴Cn ＜ Dn/7
即7Cn ＜ Dn
设数列{Dn}的前n项和为 Sn,
则有 Sn = 4[1 - (1/2)^n] /(1 - 1/2)
= 8 - (1/2)^(n-3)
当n→∞时,{Dn}的无穷项和S = limSn = lim[ 8 - (1/2)^(n-3) ] = 8
所以有,{Cn}的前n项和Tn满足
7Tn = 7C1+7C2+7C3+7C4+7C5 + 7C6+……+7Cn
＜ (7C1+7C2+7C3+7C4+7C5 ) + (D6+…… + Dn)
= (7C1+7C2+7C3+7C4+7C5 ) + S - S5
= (7/4 + 7/7 + 7/13 + 7/25 + 7/49) + 8 - [ 8 - (1/2)^(5-3) ]
= 7/4 + 7/7 + 7/13 + 7/25 + 7/49 + 1/4
≈ 3.9613
＜ 4
即,Tn＜4/7 ≈ 0.5714
∴当n≥6时,Tn＜4/7成立
而,当n=1时,T1 = 1/4 ＜4/7
当n=2时,T2 = 1/7 + T1 ≈ 0.3929 ＜4/7
当n=3时,T3 = 1/13 + T2 ≈ 0.4698 ＜4/7
当n=4时,T4 = 1/25 + T3 ≈ 0.5098 ＜4/7
当n=5时,T5 = 1/49 + T4 ≈ 0.5302 ＜4/7
综上所述,对任意n∈N ,Tn< 4/7恒成立