(10.5) Complete the proof of Theorem 10.4 by showing that if is an unbounded decreasing sequence, that .
Proof. Let be an unbounded increasing sequence. Let . Since the set is unbounded and it is bounded above by , it must be unbounded below. Hence for some we have . Also implies , so
Let be a sequence defined by and . Show that converges to a real number. You may use that if then . (Hint: Use the Monotone Convergence Theorem.)
Proof. We first prove by induction that for all . The base case is from that, . Suppose now that . Then,
Notice that this implies then that . Hence for all . This means that is monotone increasing and bounded, so the Monotone Convergence Theorem for sequences says that it must converge to some real number. .
(10.8) Let be an increasing sequence of positive numbers. Let be defined by,
Prove that is in increasing sequence. (The are called Cesaro means)
Proof. We will show that by showing that is nonnegative.
As is positive, this fraction is nonnegative if its numerator is. So, we examine:
As each of the is nonnegative as for each (as is positive and increasing) the whole sum is nonnegative. So the numerator is nonnegative, and we conclude that is nonnegative, so , so is an increasing sequence.
For each of the following sequences, find the and the .
(11.8) Prove that for any sequence . You may use the result of exercise 5.4 without proving it.
Proof. We have that :
It remains to show the result of exercise 9.10(b) which says that if and only if . Suppose that . Suppose that . Then there exists so that for all , . But this means . So for all as was arbitrary. The reverse direction follows by a similar argument.