由等差数列求和,S[n] = n(a[1]+a[n])/2 = n(3+2n+1)/2 = n(n+2).
1/S[n] = 1/(n(n+2)) = 1/2·(1/n-1/(n+2)).
T[n] = 1/2·(1-1/3)+1/2·(1/2-1/4)+...+1/2·(1/n-1/(n+2)) = 1/2+1/4-1/(2n+2)-1/(2n+4) < 3/4.
而S[n] ≥ 0,故T[n]单调递增,T[n] ≥ T[1] = 1/S[1] = 1/a[1] = 1/3.