Existence of a particular function Given positive integers $n,d,\ell\ge1$ such that $n\ge 1+(d-1)\ell$, I must find a a function
$$
a=a(n,d,\ell),
$$
such that
\begin{align*}
0< a(n,d,\ell)&<1,\\
\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)&\le a(n,d,\ell),
\end{align*}
for all $n,d,\ell\ge1,\ n\ge1+(d-1)\ell.$
I've been able to solve only the case $\ell=1$. In such a case, the function
$$
a(n,d,1)=\frac{n-d}{n-d+1}
$$
satisfy the requests.
Any help is appreciated.
 A: Consider the function $a(n, d, l) = \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n+1)} {\Gamma(d+1)}$ where $b(c, l) = \Gamma(3+ \frac {c \cdot \ell} {\ell-1})$. This function satisifies the constraint since:
\begin{align*}
\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)&=\frac{n}{n-(d-1)(\ell-1)} \cdot \frac {1} {b(n-1-(d-1), \ell)} \cdot \frac {\Gamma(n)} {\Gamma(d)} \\
&\le \frac{n}{1 + (d-1)\ell-(d-1)(\ell-1)} \cdot \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n)} {\Gamma(d)} \ \textit{(Since $n\ge 1+(d-1)\ell$)}\\
&\le \frac{n}{1 + (d-1)} \cdot \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n)} {\Gamma(d)} \\
&\le \frac{n}{d} \cdot \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n)} {\Gamma(d)} \\
&\le \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n+1)} {\Gamma(d+1)} \\
&\le a(n, d, l)
\end{align*}
Clearly $a(n,d,l)>0$ since gamma function is always positive for positive argument. To prove $a(n,d,l)<1$, let us prove the following claim:
\begin{equation*}
n < 2 + \frac {(n-d) \cdot \ell} {\ell-1}
\end{equation*}
Proof of claim:
\begin{align*}
n &\ge 1 + (d-1)\ell \\
n-n\ell &\ge 1 + d\ell -\ell -n\ell \\
n(1-\ell) &\ge 1-\ell - (n-d)\ell \\
n &\le 1- \frac {(n-d)\ell} {1-\ell} \\
n &\le 1+ \frac {(n-d)\ell} {\ell-1} \\
\implies n &< 2 + \frac {(n-d)\ell} {\ell-1}
\end{align*}
From the above claim,we get $n +1 < 3 + \frac {(n-d)\ell} {\ell-1}$ which implies that $\frac {\Gamma(n+1)} {b(n-d, l)} = \frac {\Gamma(n+1)} {\Gamma(3 + \frac {(n-d) \cdot \ell} {\ell-1})}  < 1$. Hence $a(n, d, l) = \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n+1)} {\Gamma(d+1)} < 1$
