<s>[INST] Looking for analytical proof that this function given as a power series is constant. Answering a recent question I came across the following function ($t$ is a positive integer), defined for $0\le x\le 1$ as:
$$
P_t(x)=x^t\sum_{n=0}^\infty \frac1{n+1}\binom{t(n+1)}n\left[(1-x)x^{t-1}\right]^n.
$$
Is there an analytical way to prove that $P_t(x)$ is a continuous function on $x\in(0,1)$ and:

*

*$P_t(x)=1$ for all $x:\ 0\le 1-x\le\frac1t$,

*$P_t(x)<1$ for all $x:\ \frac1t< 1-x\le1$.

I proved this for trivial cases $t=1,2$ but did not find a way to deal with general $t$.
The behavior of the function for $t=1,2,3,4$ is demonstrated below:

 [/INST] Too long for comments.
Assuming that $t$ is a positive integer, there is no problem to get explicit formulae up to $t=3$.
For $t \geq 4$ appear hypergeometric function and the result write
$$P_t(x)=\frac x {1-x} \big[Q_t(x)-1\big]$$
$$Q_4(x)=\,
   _3F_2\left(\frac{1}{4},\frac{2}{4},\frac{3}{4};\frac{2}{3},\frac{4}{3};\frac{4^4}{3^3} (1-x) x^3\right)$$
$$Q_5(x)=\,
   _4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{2}{4},\frac{3}{4},\frac{5}{4};\frac{5^5}{4^4} (1-x) x^4\right)$$
$$Q_6(x)=\,
   _5F_4\left(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{6}{5};\frac{6^6} {5^5}(1-x) x^5\right)$$
$$Q_7(x)=\,
   _6F_5\left(\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7};\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6},\frac{7}{6};\frac{7^7}{6^6} (1-x) x^6\right)$$
For $x=x_*=1-\frac 1t$, $Q_t(x_*)=\frac t{t-1}$ and then $P_t(x_*)=1$ as you observed.
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