Sorry, I was editing while you were responding.
Section A
Let be nonincreasing and .
Then, we have
which indicates by induction that for all .
Clearly, exists and holds.
Passing limit as on both sides of the recursion relation, we get or equivalently .
Thus, must be true.
However, I really do not think that the given relation and the intitial condition gives .
Section B
But if you can show , the rest follows by induction since
is true if holds.
I will be thinking on it...
Section C
Set for .
Hence, it is easy to see that for all since holds for all .
Now, we shall show , which implies .
Apply Section B to obtain the result.
Section D
Therefore, is a decreasing sequence which tends to from above at .
The question asks to prove it's monotone and convergent, then to find the limit.
I've proven it's monotone and convergent; by the Monotone Convergence Theorem I can show that the lim= inf( ).
What you did in your original post isn't tools we've used yet to prove infinums of sets yet. Hence, I've got to figure out another way to do it, and every way I"ve tried has ended in failure.
First, we show that its decreasing, next we show that it is bounded from below (in worst case it is bounded below by ), this means that it has a limit at infinity.
Then in the recursion we pass to limits on both sides (note that if then as ; that, is every subsequence converges to the same limit).
Finally, we prove that , which is the desired value since is decreasing.
I know I repeat the same, but no other idea to show .