Let Sn be a sequence of nonnegative numbers and for each n defineOn = 1n S1 + S2 + + Sna Showlim inf Sn lim inf on limsup on limsup sn-b Show that if lim sn exists then limon exists and limon=lim

(a)

First, we will show that lim inf Sn < lim inf On.

Let x = lim inf Sn. This means that for any ε > 0, there exists an N such that Sn < x + ε for all n ≥ N.

Now, let's consider On. By definition,

On = 1/n (S1 + S2 + ... + Sn) ≤ 1/n (n(x + ε)) = x + ε/n.

This means that for any ε > 0 and any n, there exists an N such that On ≤ x + ε for all n ≥ N. Thus, lim inf On ≤ x.

Next, we will show that lim inf On < lim sup On.

Let y = lim sup On. This means that for any ε > 0, there exists an N such that On > y - ε for infinitely many values of n.

Now, let's consider Sn. By definition,

Sn = S(n-1) + an ≥ n-1/n y

where an = On - On-1.

Dividing by n and rearranging, we get

On ≤ y/n + (S(n-1))/n.

Taking the lim inf as n → ∞, we get

lim inf On ≤ lim sup S(n-1)/n = lim sup Sn/n.

Since lim sup Sn/n = x (by the Squeeze Theorem), we have lim inf On ≤ x.

Finally, we will show that lim sup On < lim sup Sn.

Let z = lim sup Sn. This means that for any ε > 0, there exists an N such that Sn > z - ε for infinitely many values of n.

Now, let's consider On. By definition,

On = 1/n (S1 + S2 + ... + Sn) ≥ 1/n (n(z - ε)) = z - ε.

This means that for any ε > 0, there exists an N such that On ≥ z - ε for infinitely many values of n. Thus, lim sup On ≥ z.

Putting it all together, we have

lim inf Sn < lim inf On ≤ lim sup On < lim sup Sn.

(b)

Assume that lim Sn exists. Let L = lim Sn.

We will show that lim On = L.

Let ε > 0. By definition of lim Sn, there exists an N such that Sn < L + ε/2 for all n ≥ N.

Now, consider On. By definition,

On = 1/n (S1 + S2 + ... + Sn) = (1/n)Sn + (1/n)(S(n-1)) + ... + (1/n)S1.

Since each term in the sum is nonnegative, we can use the triangle inequality to get

On ≤ (1/n)Sn + (1/n)(S(n-1) + Sn) + ... + (1/n)(S1 + S2 + ... + Sn)

= (1/n)Sn + (1/n)(S(n-1) + Sn) + ... + (1/n)(S1 + S2 + ... + S(N-1) + ... + Sn)

= (1/n)Sn + (1/n)(S(n-1) + Sn) + ... + (1/n)(S1 + S2 + ... + S(N-1)) + (1/n)Sn + ... + (1/n)Sn

= (1/n)Sn + (n-N+1)/n (1/(n-N+1))(S(N-1) + ... + Sn)

≤ (1/n)Sn + (n-N+1)/n (1/(n-N+1))(Sn + ... + Sn)

= (1/n)Sn + (n-N+1)/n (1/(n-N+1))(N-1)Sn

= (N-1)/n Sn + Sn.

Now, if n ≥ N, then

On ≤ (N-1)/n Sn + Sn < (N-1)/N (L + ε/2) + L + ε/2 = L + ε.

Thus, for any ε > 0, there exists an N such that On < L + ε for all n ≥ N. Therefore, lim On = L.

(c)

Let Sn = 1 if n is odd and Sn = 0 if n is even.

Then, On = (1 + 0 + 1 + 0 + ...) / n = 1/2 for all n. Thus, lim On = 1/2.

However, lim Sn does not exist, since the even terms converge to 0 and the odd terms converge to 1