In a), I think you could solve it by looking at the function
where
and by realizing that a k that gives a minimal or a maximal value for f, will also give a minimal or a maximal value, respectively, for F.
Also note that
I'm going to assume that I am interpreting these questions correctly.
Problem: Let . Find
Solution: There is no "minimum" of this set. Those concepts (at least for how I have them defined) act only on members of the set. So existing would imply such that . Before we proceed we need a lemma.
Lemma: Let , then
Proof: This is just true by a series of observations. Namely:
. Adding to both sides gives . Dividing both sides by gives The conclusion follows.
Since is an increasing function this implies that is a decreasing function.
Thus we can show that does not exist. We do this by contradiction. Suppose . Then for some . But and contradicting our choice of . Therefore does not exist.
From our lemma we can easily deduce that and . Thus, . Consequently,
Lastly we must find . We claim that it is.......I leave the rest for you.
Well, you should have learned to at least calculate some of those numbers!
In the first problem, you have numbers of the form .
When k= 1, that is .
When k= 2, that is .
When k= 3, that is .
That is, so your sequence reduces to . It should be easy to see that that is a decreasing sequence.