Let Sn be a sequence of nonnegative numbers and for each n defineOn = 1n S1 + S2 + + Sn Showlim inf Sn lim inf on limsup on limsup sn

First, we can show that lim inf Sn ≤ lim inf On ≤ lim sup On ≤ lim sup Sn:

• lim inf Sn ≤ lim inf On: Since On is an average of a finite number of terms in Sn, we have On ≤ Sn for all n. Thus, if Sn converges to some limit L, then On must also converge to L (since it is sandwiched between Sn and a constant sequence with limit L). Therefore, lim inf Sn ≤ lim inf On.
• lim inf On ≤ lim sup On: This follows from the definition of lim inf and lim sup. Any subsequence of On has a further subsequence that converges to lim inf On, and any subsequence of this further subsequence has a further subsequence that converges to lim sup On. Therefore, lim inf On ≤ lim sup On.
• lim sup On ≤ lim sup Sn: Similar to the first point, we have Sn ≤ nOn for all n, so if Sn converges to some limit L, then On must also converge to L (since it is sandwiched between Sn and a sequence that converges to L). Therefore, lim sup On ≤ lim sup Sn.

Now, to prove the original statement, we use the following lemma:

Lemma: If a sequence (an) converges to L, then any subsequence (an_k) has a lim inf and lim sup that both converge to L.

Proof: Let (an_k) be a subsequence of (an). Then, since (an) converges to L, we have for any ε > 0, there exists N such that n ≥ N implies |an - L| < ε. Since (an_k) is a subsequence, we can choose N so that k ≥ N implies n_k ≥ N, and thus |an_k - L| < ε for all k ≥ N. Therefore, lim inf an_k = lim sup an_k = L.

Using this lemma, we can prove the original statement:

• lim inf Sn < lim inf On: Let (Sn_k) be a subsequence of (Sn) that converges to lim inf Sn. By the lemma, we know that lim inf On ≤ lim inf Sn_k for any subsequence (On_k) of (On). In particular, we can choose On_k to be the subsequence of On consisting of the first k terms. Then, we have On_k = 1/k (S1 + S2 + ... + Sk), so lim inf On_k = lim inf On by continuity of multiplication. But On_k converges to lim inf On (since it is a subsequence), so we must have lim inf On ≤ lim inf Sn_k. Taking the lim inf over all subsequences (On_k), we get lim inf On ≤ lim inf Sn.
• lim inf On < lim sup On: This was shown earlier.
• lim sup On < lim sup Sn: Let (Sn_k) be a subsequence of (Sn) that converges to lim sup Sn. By the lemma, we know that lim sup Sn_k = lim inf Sn_k = lim inf On_k ≤ lim sup On_k for any subsequence (On_k) of (On). In particular, we can choose On_k to be the subsequence of On consisting of the last k terms. Then, we have On_k = 1/k (Sn - S(n-k)), so lim sup On_k = lim sup On by continuity of multiplication. But On_k converges to lim sup On (since it is a subsequence), so we must have lim inf Sn_k ≤ lim sup On. Taking the lim sup over all subsequences (Sn_k), we get lim sup Sn ≤ lim sup On