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I encounter the following recurrence \begin{equation} \tag{1}\label{eq:1} h_{j+1} = h_{j} ( 1 - 1/j - c h_j ), \quad j \geq j_0, \end{equation} with $h_{j_0}>0$, $c>0$ and $0< 1-1/j_0 - c h_{j_0} < 1$.

Question: How does $h_j$ behave for large $j$?

I believe that $$\tag{2} \label{eq:3} h_j = O(1/j^{1 + \alpha}),$$ for some $\alpha>0$ dependent on $c$, and it seems like I have half a proof:

Lemma: If $$\tag{3} \label{eq:2} x_{j+1} \leq x_{j} ( 1 - \beta/j), \quad j \geq j_0,$$ with $\beta>0$, $x_{j_0}>0$, $0< 1-\beta/x_{j_0}< 1$, then $x_j \leq (x_{j_0}/j_0) j^{-\beta}$.

Proof: We have \begin{align} x_{N+1} &= x_{j_0} \prod_{j=j_0}^N (1-\beta/j) = x_{j_0} \exp\left(\sum_{j=j_0}^N \log(1-\beta/j) \right) \leq x_{j_0} \exp\left(-\sum_{j=j_0}^N \beta/j \right) \\ &\leq x_{j_0} \exp\left( -\beta \int_{j_0}^{N+1}1/x \,dx \right) = x_{j_0} \exp\left(-\beta (\log(N+1)-\log(j_0)) \right) \\ = &(x_{j_0}/j_0) (N+1)^{-\beta}. \quad \quad \mbox{Q.E.D.} \end{align}

In above, we use the inequality $\log(1+x)\leq x$, which is, of course, tight for small $x$. In particular, for any small $\epsilon>0$, we have the inverse inequality $(1-\epsilon) x \leq \log(1+x)$. Therefore, we can also get the following:

Lemma: If $x_{j+1} \geq x_{j} ( 1 - \beta/j)$, $j \geq j_0$, with $\beta>0$, $x_{j_0}>0$, $0< 1-\beta/x_{j_0}< 1$, then for any $\epsilon>0$, there is a $C_\epsilon>0$ such that $x_j \geq C_\epsilon j^{-\beta(1-\epsilon)}$.

To get something out of \eqref{eq:1}, we can use a bootstrapping strategy. First, we have $$ h_{j+1} < h_{j} ( 1 - 1/j ), \quad j \geq j_0. $$ So, by the first lemma, we have $h_j < (h_{j_0}/j_0) j^{-1}$.

The $O(1/j)$ asymptotic for \eqref{eq:1} cannot be tight, otherwise $h_j \geq \gamma / j$ for some $\gamma>0$. But then, by \eqref{eq:1}, $h_{j+1} \leq h_j(1-1/j-\gamma/j) = h_j( 1- (1+\gamma)/j)$, which, by the first lemma, implies that $h_j = O(1/j^{1+\gamma})$, causing a contradiction.

Thus $h_j$ must decay faster than $O(1/j)$. But I don't quite have a proof that the conjecture in \eqref{eq:3} is true.

We can also get a lower bound for $h_j$ from the earlier upper bound: $h_{j+1}> h_j(1-1/j-(h_{j_0}/j_0)(1/j) = h_j(1 - (1+h_{j_0}/j_0)/j)$. This shows, by the second lemma above, $h_j$ cannot decay faster than $O(1/j^{(1+h_{j_0}/j_0)})$.

This is about as much bootstrapping as I can do. If anyone has seen anything related, or know of a proof to \eqref{eq:3} and the optimal value for $\alpha$ thereof, I will be thrilled to hear from you. Thanks.

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    $\begingroup$ The actual answer is $h_j=\frac{1+o(1)}{cj\log j}$ as $j\to\infty$. $\endgroup$
    – fedja
    May 13 '19 at 9:13
  • $\begingroup$ Thanks very much. Sorry for the late response. I was wondering if this follows from a reasonably well-known trick/method... I also wonder if your 'c' is my 'c' in (1). $\endgroup$
    – Thomas
    May 21 '19 at 16:31
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    $\begingroup$ Yes, it is.$\hspace{5pt}$ $\endgroup$
    – fedja
    May 21 '19 at 17:00
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It looks that it is not so, right behaviour is $(n\log n)^{-1}$.

Clearly your sequence is positive and decreasing. Denote $h_j=1/x_j$. Then $$x_{j+1}=x_j(1-1/j-c/x_j)^{-1}\geqslant x_j(1+(1/j+c/x_j))=x_j+c+x_j/j.$$

Further denoting $x_{j}=j y_j$ we rewrite this as $y_{j+1}\geqslant y_j+c/(j+1)$. Summing up it for $j=j_0,\ldots,n-1$ it yields $y_n\geqslant c\log(n)-C$ for certain constant $C>0$, that is $x_n\geqslant cn\log(n)+O(n)$. It means that $$\log x_{j+1}-\log x_{j}=-\log\left(1-\frac1j-\frac{c}{x_j}\right)\leqslant \frac{1}j+\frac{1}{j\log j}+O\left(\frac{1}{j\log^2 j}\right), $$ thus $\log x_n\leqslant \log n+\log\log n+O(1)$, so $x_n\leqslant {\rm const}\cdot n\log n$.

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