Define the following recurrence for $n = 1, 2, \cdots$
$T(n) = ( 1 - \operatorname{H}(\frac{1 - P^{\frac{1}{n}}}{2}))^n$ where $0 < P < 1$ is a constant, function $\operatorname{H}(\cdot)$ is the binary entropy function: $\operatorname{H}(p) \triangleq -p\log_2 p - (1-p)\log_2 (1-p)$.
Numerically, I tried many examples for different P and n, and all examples suggest that the recurrence be monotonically decreasing, that is:
$T(n) > T(n-1)$ for all $n = 2, 3, \cdots$
Is there an analytical way to prove or disprove my conjecture? Any suggestion is appreciated.
I have an idea to approach this, but the algebra got a bit messy towards the end so I didn't quite finish. I will say it anyway and then you can figure out if it works.
First, some changes in notation which I think helps. Forget about doing it for natural numbers and let $x$ be a real number playing the role of $1/n$ in your example. Let $p=e^{a}$ and $$g(x)=1-H\left(\frac{1-e^{ax}}{2}\right).$$ We can rephrase your question as to whether $S(x)=\log T(1/x)$ is decreasing or increasing (not sure which one you want from the way you asked, but it doesn't matter), where $$S(x)=\frac{1}{x} \log_2 g(x).$$ Now take the derivative and check whether you can show that it is positive or negative.