The reciprocal n/an = (2a (n-1)+n-1)/3a (n-1) = 2/3+(n-1)/3a (n-1).

Let n/an=bn, bn= 1/3b(n- 1)+2/3.

That is bn-1=1/3 (b (n-1)-1).

So {bn- 1} is a geometric series, and the first term is b1-1=1-1/3.

So bn- 1 =- 1/3 n

bn= 1-3^n

an=n/bn=n*3^n/(3^n- 1)

(2) It is proved that 3/(3-1) * 3 2/(3 2-1) * 3/(3 3-1) * ... * 3N/(3N-65438.

It is proved that when n∈N* exists (1-1/3) (1-kloc-0//3 2) ... =1-(1)

The following is proved by mathematical induction.

When n= 1, ① holds.

Assume that n=k holds, that is,

( 1- 1/3)( 1- 1/3^2)...( 1- 1/3^k)>； = 1-( 1/3+ 1/3^2+... 1/3^k)

Then when n=k+ 1

( 1- 1/3)( 1- 1/3^2)...( 1- 1/3^k)[ 1- 1/3^(k+ 1)]>； = 1-( 1/3+ 1/3^2+... 1/3^k)[ 1- 1/3^(k+ 1)]

= 1-( 1/3+ 1/3^2+... 1/3^k)- 1/3^(k+ 1)+ 1/3^(k+ 1)( 1/3+ 1/3^2+... 1/3^k)

& gt =1-[1/3+1/32+...1/3k+1/3 (k+1)] ① formula holds.

Therefore, from the mathematical induction, formula ① holds for all n∈N*.

∴( 1- 1/3)( 1- 1/3^2)...( 1- 1/3^n)>； = 1-( 1/3+ 1/3^2+... 1/3^n)

= 1-( 1/3)[ 1-( 1/3)^n]/( 1- 1/3)

= 1-( 1/2)[ 1-( 1/3)^n]

= 1/2+ 1/2( 1/3)^n

& gt 1/2

That is, the original formula holds, and 2, the known sequence {an} satisfies: a 1=3/2, an = (3 an-1)/(2an-1+n-1) (n ≥ 2, n ?.

(2) Prove the inequality A 1 A2 A3 An.