I have a solution for the following problem but I really wonder what kind of ways will be provided for this.
Problem. Let
Prove that
where
.................
The sequence is "almost" constant. Let's bound from below by replacing by in the first term and by in the second one (after the minus sign): we get . However, we have so that this lower bound is slightly too strong, but if we had a positive factor instead in front of (no matter how small), this would be right. Then, to bound the first term, it suffices to consider the indices where and (you can bound the terms with other indices by 0 thanks to the previous computation). There are about such couples , and for each of them the summand is greater than , which is strictly greater than . We get that for some positive . Choosing for instance, you easily see that this is unbounded from above.
Here is my solution.
Solution. Clearly, we have and for all .
Therefore, we have
........(1)
and
.................................................. ...............(2)
for all .
Therefore, using (1) and (2), we have the following inequalities:
......(3)
.................................................. .........._(4)
.................................................. ..........,(5)
for all .
Hence, from (3)--(5), we learn that
holds for all , where
for .
It suffices to prove that is unbounded.
As , we have
.
Thus, the proof is done.
Notation. denotes the least integer function and denotes the greatest integer function.
Appendix. Here, (2) is shown.
By the periodicity, we obtain
.................................................. .......................(6)
for all .
Using (6), we get
......................................_
......................................_
for all .