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.



Nonlinear recurrence
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 $x\_n\leqslant {\rm const}\cdot n\log n$.


