Let the probability $p\in\left(0,1\right)$ be defined through the equation:
$\left\{ 1-\left(1-p\right)^{n}\right\} \cdot\frac{1}{n\cdot p}=c$
where $c\in\left(\frac{1}{n},1\right)$ and $n$ is a natural number. I wish to prove that $p=p\left(n\right)$ is decreasing and that $n\cdot p$ is also decreasing in $n$. Thank you for any inputs!
-- UPDATE --
Showing that $p=p\left(n\right)$ is decreasing can be seen by taking the derivative indeed. For the behavior of $\lambda(n):=n\cdot p(n)$ notice the following. Let $R\left(\lambda,n\right)=\left\{ 1-\left(1-\frac{\lambda}{n}\right)^{n}\right\} \frac{1}{\lambda}-c$. By the Implicit Function Theorem applied to $R\left(\lambda,n\right)=0$ we have
$ \frac{d\lambda}{dn}=-\frac{\frac{\partial R}{\partial n}}{\frac{\partial R}{\partial\lambda}}<0 $
provided that the partial derivatives satisfy my conjecture: $\frac{\partial R}{\partial n}<0$ and $\frac{\partial R}{\partial\lambda}<0$. How to show this last part?
In order to prove $p$ is decreasing in $n$, you could differentiate the given equation w.r.t $n$,
$$-(1-(1-p)^n)\dfrac{1}{(np)^2}\,(p + n\,\dfrac{dp}{dn}) + \dfrac{1}{np}\,(1-p)^n\,log(1-p)\,\dfrac{dp}{dn} = \dfrac{dc}{dn}$$
Rearranging terms,
$$\dfrac{dp}{dn}\,\left(\dfrac{-1}{np^2}\,(1-(1-p)^n)\, + \dfrac{(1-p)^n\,log(1-p)}{p\,n} \right) = \dfrac{dc}{dn} + \dfrac{1-(1-p)^n}{p\,n^2}$$
$$\dfrac{dp}{dn} \left(\dfrac{(1-p^n)(1+p \, log(1-p))-1}{n\,p^2}\right) = \dfrac{dc}{dn} + \dfrac{1-(1-p)^n}{p\,n^2}$$
Since $1/n<c<1$ for each $n$, $c$ is increasing in $n$. Hence RHS $\geq0$. Moreover, $(1-p^n)(1+p \, log(1-p)) < 1$ since $0<p<1$. Hence, $\dfrac{dp}{dn}< 0$ ; thus $p$ is decreasing in $n$.
Let $E(p,n) = p\,n$ and $F(p) = (1-p^n)(1+p \, log(1-p))-1 <0$. Differentiating $E$ w.r.t $n$, $\dfrac{dE}{dn} = p + n\,\dfrac{dp}{dn}$. From previous result,
$$p + n\,\dfrac{dp}{dn} = p + \left(\dfrac{n^2\,p^2\dfrac{dc}{dn}}{F(p)} + \dfrac{p(1-(1-p)^n)}{F(p)}\right)$$ $$= \dfrac{n^2\,p^2\dfrac{dc}{dn}}{F(p)} + p \left(\dfrac{1-(1-p)^n + F(p)}{F(p)}\right) $$ $$= \dfrac{n^2\,p^2\dfrac{dc}{dn}}{F(p)} + p \left(\dfrac{-(1-p)^n + (1-p)^n (1+p\,log(1-p))}{F(p)}\right) $$ $$= \dfrac{n^2\,p^2\dfrac{dc}{dn}}{F(p)} + p \left(\dfrac{(1-p)^n\,p\,log(1-p)}{F(p)}\right) $$
Now, $log (1-p) <0$; so is $F(p)$. The first expression is $< 0$ but the second one is non-negative so it's difficult to argue that $dE/dn < 0$. (Let me check this later)