1/(n-i+1), upon mapping the last element of the current cycle to the one which begins it As the total number Kn(T)of cycles of T is exactly the number of times an element closes the loop upon completing its own cycle Kn(丌)=X1+…X a sum of independent, but not identically distributed random variables Hence, despite the similarity of(5) to(1), the hypotheses of the classical central limit theorem do not hold. Nevertheless, in 1922 Lindeberg 7 pro- vided a general condition which can be applied in this case to show that Kn(T) is asymptotically normal To explore Lindeberg's condition, first consider the proper standardiza- tion of Kn(T) in our example. As any Bernoulli random variable with success probability p has mean p and variance p(1-p), the Bernoulli variable Xi in (4) has mean i-I and variance i-(1-i-l)for i=1,., n. Thus, 1 hn=∑;and ∑ mean and variance of Kn(T), respectively; the mean hn is known as harmonic number. In particular, standardizing Kn(r) to have mean zero and variance 1 results in Kn(丌)-hn which, absorbing the scaling inside the sum, can be written as Wn=∑ Xin where Xi In general, it is both more convenient and more encompassing to deal not with a sequence of variables but rather with a triangular array as in which satisfies the following condition Condition 1.1 For every n= 1, 2,.. the random variables making up the collection Xn=(Xi n: Isisn are independent with mean zero and finite variances a n= Var(Xi n), standardized so that Xi,n has variance Var(Wn)1/(n − i + 1), upon mapping the last element of the current cycle to the one which begins it. As the total number Kn(π) of cycles of π is exactly the number of times an element closes the loop upon completing its own cycle, Kn(π) = X1 + · · · Xn, (5) a sum of independent, but not identically distributed random variables. Hence, despite the similarity of (5) to (1), the hypotheses of the classical central limit theorem do not hold. Nevertheless, in 1922 Lindeberg [7] pro￾vided a general condition which can be applied in this case to show that Kn(π) is asymptotically normal. To explore Lindeberg’s condition, first consider the proper standardiza￾tion of Kn(π) in our example. As any Bernoulli random variable with success probability p has mean p and variance p(1 − p), the Bernoulli variable Xi in (4) has mean i −1 and variance i −1 (1 − i −1 ) for i = 1, . . . , n. Thus, hn = Xn i=1 1 i and σ 2 n = Xn i=1  1 i − 1 i 2  (6) are the mean and variance of Kn(π), respectively; the mean hn is known as the n th harmonic number. In particular, standardizing Kn(π) to have mean zero and variance 1 results in Wn = Kn(π) − hn σn , which, absorbing the scaling inside the sum, can be written as Wn = Xn i=1 Xi,n where Xi,n = Xi − i −1 σn . (7) In general, it is both more convenient and more encompassing to deal not with a sequence of variables but rather with a triangular array as in (7) which satisfies the following condition. Condition 1.1 For every n = 1, 2, . . ., the random variables making up the collection Xn = {Xi,n : 1 ≤ i ≤ n} are independent with mean zero and finite variances σ 2 i,n = Var(Xi,n), standardized so that Wn = Xn i=1 Xi,n has variance Var(Wn) = Xn i=1 σ 2 i,n = 1. 4