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

.................

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- Oct 4th 2008, 02:50 AMbkarpuzUnbounded sequence?
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

................. - Oct 4th 2008, 04:04 AMLaurent
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.

- Oct 4th 2008, 04:35 AMbkarpuz
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 .