Consider a strictly positive sequence $\{u_n\}$ of real numbers (i.e. $\forall n, u_n >0$). Suppose that $\displaystyle \lim_{n \to \infty} u_n = 0$ and that $\{u_n\}$ is not decreasing.
Consider the sequence $\{l_n\}$ defined by : $\forall n \in \mathbb N, l_n := \min\left\{i-n,\text{ such that } u_i\leq u_n,\forall i> n\right\}$.
This sequence counts the number of iterations needed before a term of the sequence $\{u_n\}$, smaller than the current term, appears. My question is whether this sequence is bounded.
If $\{u_n\}$ is a decreasing sequence, it is clear that $\forall n \in \mathbb N, l_n = 1$, so in this case, the sequence is bounded.
But my question concerns the case where $\{u_n\}$ is not decreasing !
Given that the sequence $\{u_n\}$ converges towards $0$, even if it is not a decreasing sequence, the sequence $\{l_n\}$ is well defined.
In fact, given an iterate $u_n$, the convergence to $0$ of the sequence can be written as : \begin{equation}(*)\text{ there exists } i(n) \in \mathbb N, \text{ such that } \forall k \geq i(n), u_k \leq u_n\end{equation}
So we just have to take $l_n = \min\left\{i(n)-n, \text{where } i(n) \text{ verifies }(*)\right\}$.
As we just see, $\{l_n\}$ is a (non-negative) well defined sequence. But is $\{l_n\}$ bounded?
I tried to think of a counter-example, but I have the impression that this result looks true (I mean, that the sequence is bounded).
However, I don't know how to prove it. Do you have any clues?
Best Answer
$$.10,\;.11,$$ $$.0100,\;.0101,\;.0110,\;.0111,$$ $$.001000,\;.001001,\;.001010,\;.001011,\;.001100,\;.001101,\;.001110,\;.001111,$$ $$.00010000,\;.00010001,\;.00010010,\;.00010011,\;.00010100,\;.00010101,\;.00010110,\;.00010111, \\ .00011000,\;.00011001,\;.00011010,\;.00011011,\;.00011100,\;.00011101,\;.00011110,\;.00011111,$$ $$\cdots$$