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**Opalg** The answer is No. In other words, such a sequence $\displaystyle (y_n)$ does not necessarily eventually decrease.

To start with, consider the sequence $\displaystyle (x_n)$ defined by $\displaystyle x_n = \tfrac k{k+1}$ for $\displaystyle k^2+1\leqslant n\leqslant (k+1)^2\ (k=0,1,2,\ldots).$ Then

$\displaystyle y_{k^2+1} = \frac{x_{k^2+1}}{k^2+1} = \frac k{(k+1)(k^2+1)},\qquad y_{k^2} = \frac{x_{k^2}}{k^2} = \frac {k-1}{k^3},$

and you can check by multiplying out the fractions that $\displaystyle y_{k^2+1} > y_{k^2}$ for all k. Thus there are infinitely many values of n for which $\displaystyle y_{n+1}>y_n.$

That example does not quite answer the original question, because the sequence $\displaystyle (x_n)$ is not strictly increasing (being constant throughout the interval $\displaystyle k^2+1\leqslant n\leqslant (k+1)^2$). However, in principle there is no difficulty in adjusting the sequence so as to make it strictly increasing. For each k, you can make $\displaystyle x_{k^2+1}$ very slightly smaller, without disturbing the property that $\displaystyle y_{k^2+1} > y_{k^2}$. Then you can interpolate the values of $\displaystyle x_n$ linearly for $\displaystyle k^2+1\leqslant n\leqslant (k+1)^2$ so as to ensure that the sequence increases strictly.