Let Sn be a sequence of nonnegative numbers and for each n defineOn = = 81 + 82 + + Sna ShowequenceNslim inf Sn lim inf on limsup on limsup sn-Hint For the last inequality show first that M N imp

(a) We first note that since each term in the sequence (On) is a sum of nonnegative numbers, it is increasing. Thus, if lim inf Sn exists, then lim inf On exists and is equal to lim Sn, as each term in (On) is bounded below by Sn. Similarly, if lim sup Sn exists, then lim sup On exists and is equal to lim Sn, as each term in (On) is bounded above by Sn.

To show that lim inf Sn < lim inf On, we observe that for any n, we have

On = (81 + 82 + ... + Sn) ≥ Sn,

since each term in the sum is nonnegative. Thus, lim inf On ≥ lim inf Sn.

To show that lim inf On < lim sup On, we note that since (On) is increasing, it is either convergent or unbounded. If it is convergent, then lim inf On = lim sup On, so the inequality holds. If it is unbounded, then for any M, there exists an N such that On > M for all n > N. But then

sup(on: n > N) ≥ M,

since each term in the supremum is at least M. Thus, lim sup On = ∞, and the inequality holds.

To show that lim sup On < lim sup Sn, we use the hint. Let M < N be given. Then for any n > M, we have

sn = On − (81 + 82 + ... + (n − 1)) ≤ On − (81 + 82 + ... + N) ≤ sup(on: n > N),

since each term in the sum is nonnegative. Taking the lim sup as n → ∞, we obtain

lim sup sn ≤ sup(on: n > N).

Taking the lim sup as N → ∞ and using the fact that lim sup On ≤ lim sup Sn, we obtain

lim sup sn ≤ lim sup On.

Since M < N were arbitrary, we have

lim sup sn ≤ lim sup On ≤ lim sup Sn.

(b) Suppose lim sn = L. Then for any ε > 0, there exists N such that sn − L < ε/2 for all n > N. Let M > N be given. Then for any n > M, we have

|on − L| = |(81 + 82 + ... + sn) − (81 + 82 + ... + sn−1) − (sn − L)|

= |sn + (81 + 82 + ... + sn−1) − L − sn + L| = |(81 + 82 + ... + sn−1) − L| ≤ sn − L < ε/2.

Thus, we have shown that for any ε > 0, there exists M > N such that |on − L| < ε for all n > M. This shows that lim on = L = lim sn.

(c) Let Sn = 1 if n is odd and Sn = 0 if n is even. Then On = n/2 if n is even and On = (n + 1)/2 if n is odd. Thus, lim On = lim inf On = lim sup On = 1/2, but lim Sn does not exist