Prove the Inequality: lim inf sn < lim inf on < limsup on < limsup sn

First, we prove that lim inf sn < lim inf on < limsup on < limsup sn.

Since sn is a nonnegative sequence, we have lim inf sn = inf{n: sn} and limsup sn = sup{n: sn}. Similarly, we have lim inf on = inf{n: on} and limsup on = sup{n: on}.

Suppose that lim inf sn = L. Then, for any ε > 0, there exists N such that sn < L + ε for all n > N. This implies that on = (1/n)(s1 + s2 + ... + sn) < (1/n)(s1 + s2 + ... + sN) + (1/n)(n-N)(L+ε) = (1/n)(s1 + s2 + ... + sN) + (L+ε)(1 - N/n) Since lim(1 - N/n) = 1, we have lim inf on ≤ lim inf {(1/n)(s1 + s2 + ... + sN) + (L+ε)(1 - N/n)} = (1/N)(s1 + s2 + ... + sN) + L+ε for all ε > 0.

Taking ε → 0, we obtain lim inf on ≤ (1/N)(s1 + s2 + ... + sN) + L.

Since N is arbitrary, we have lim inf on ≤ L.

Similarly, we can show that limsup on ≥ L.

Therefore, we have lim inf sn < lim inf on ≤ limsup on ≤ limsup sn.

Next, we prove the hint.

Suppose that M > N. We have sup{on: n > M} = sup{(1/n)(s1 + s2 + ... + sn): n > M} = sup{(1/n)(s1 + s2 + ... + sN + sN+1 + ... + sn): n > M} = sup{(1/n)(s1 + s2 + ... + sN) + (1/n)(sN+1 + ... + sn): n > M} ≤ sup{(1/n)(s1 + s2 + ... + sN)} + sup{(1/n)(sN+1 + ... + sn): n > M} = (1/M)(s1 + s2 + ... + sN) + sup{(1/n)(sN+1 + ... + sn): n > N} ≤ (1/M)(s1 + s2 + ... + sN) + sup{sn: n > N} Therefore, we have sup{on: n > M} < (1/M)(s1 + s2 + ... + sN) + sup{sn: n > N}.