Show that $\sum_{k=0}^N\binom{N}{k}x^k(1-x)^{N-k}(1-\alpha k/N)^{-1}$ is decreasing in $N$ I've come across the following sum of binomial coefficients:
$$S = \sum_{k = 0}^N \binom{N}{k}x^k (1-x)^{N-k} \frac{1}{1 - \alpha \frac{k}{N}},$$
where $x \in [0, 1]$ and $\alpha \in [0, 1)$.
I need to show $S$ is decreasing in $N$ (as the last term becomes smaller when $N$ increases), regardless of values of $x$ and $\alpha$ within the range specified above (that's what I obtained from simulation). I understand that $S$ ranges between $1$ and $\frac{1}{1 - \alpha}$ but don't know how to proceed further.
Much appreciated.
 A: It's sufficient to consider $0<x,\alpha<1$.
\begin{align*}
S&=\sum_{k=0}^N\binom{N}{k}x^k(1-x)^{N-k}\int_0^1 t^{-\alpha k/N}\,dt
\\&=\int_0^1\sum_{k=0}^N\binom{N}{k}(xt^{-\alpha/N})^k(1-x)^{N-k}\,dt
\\&=\int_0^1(1-x+xt^{-\alpha/N})^N\,dt=\int_0^1 t^{-\alpha f(x,t^{-\alpha/N}-1)}\,dt,
\end{align*}
where $f(x,y)=\log(1+xy)/\log(1+y)$ for $y>0$.
So we're left to show that $f(x,y)$ increases with $y$.
Here I'm going the "straightforward calculus" way. Denoting $f_y:=\partial f/\partial y$, $$\frac{f_y(x,y)}{f(x,y)}=\frac{x}{(1+xy)\log(1+xy)}-\frac{1}{(1+y)\log(1+y)}=\frac{g(xy)-g(y)}{y},$$ where $g(z)=\dfrac{z}{(1+z)\log(1+z)}$ is decreasing (for $z>0$), because $$\frac{g'(z)}{g(z)}=\frac{1}{1+z}\left(\frac1z-\frac1{\log(1+z)}\right)<0$$ since $0<\log(1+z)<z$. Hence $g(xy)>g(y)$, thus $f_y>0$ and we're done.
A: (For comparison purposes, this was the direction I was considering...  It does not seem to provide a satisfactory conclusion.)
If $\alpha=0$ the conjecture is false.  If $\alpha=1$ the last term in the sum is indeterminate.  So we assume that $0\lt\alpha \lt 1$ and choose $N\gt \frac 1{1-\alpha}$ and note that this series is finite and geometric series with $|r|\lt 1$ are absolutely convergent, so we can write
$$\frac 1{1-\alpha\frac kN}=\sum_{i=0}^\infty\left(\alpha\frac kN\right)^i\\
\frac 1{1-\alpha\frac k{N-1}}-\frac 1{1-\alpha\frac {k+1}N}=\frac{\alpha\frac k{N-1}-\alpha\frac {k+1}N}{(1-\alpha\frac k{N-1})(1-\alpha\frac {k+1}N)}\\
=\alpha\frac{\frac {Nk-Nk-N+k+1}{N(N-1)}}{(1-\alpha\frac k{N-1})(1-\alpha\frac kN)}=\frac{\alpha}{N}\frac{\frac k{N-1}-1}{\left(1-\alpha \frac k{N-1}\right)\left(1-\alpha \frac kN\right)}$$
which can then be used as
$$S_N = \sum\limits_{k = 0}^N \binom{N}{k}x^k (1-x)^{N-k}\frac 1{1-\alpha\frac kN}=\sum\limits_{k = 0}^N \binom{N}{k}x^k (1-x)^{N-k} \sum_{i=0}^\infty\left(\alpha\frac kN\right)^i\\
=1+\sum\limits_{k = 0}^N \binom{N}{k}x^k (1-x)^{N-k} \sum_{i=1}^\infty\left(\alpha\frac kN\right)^i\qquad \text{inner sum has first term $1$}\\
=1+\sum\limits_{k = 1}^N \binom{N}{k}x^k (1-x)^{N-k} \sum_{i=1}^\infty\left(\alpha\frac kN\right)^i\qquad k=0\to \text{outer first term zero}\\
=1+\alpha\sum\limits_{k = 1}^N \binom{N}{k}\frac kNx^k (1-x)^{N-k} \sum_{i=0}^\infty\left(\alpha\frac kN\right)^i\qquad \text{divide all inner-sum terms by $\alpha\frac kN$}\\
=1+\alpha x\sum\limits_{k = 0}^{N-1} \binom{N-1}{k}x^{k} (1-x)^{N-k-1} \sum_{i=0}^\infty\left(\alpha\frac {k+1}N\right)^i\qquad \text{re-index $k$}\\
=1+\alpha x\sum\limits_{k = 0}^{N-1} \binom{N-1}{k}x^{k} (1-x)^{N-k-1} \frac 1{1-\alpha\frac{k+1}N}\qquad\text{revert inner sum}\\
=1+\alpha x\sum\limits_{k = 0}^{N-1} \binom{N-1}{k}x^{k} (1-x)^{N-k-1} \left(\frac 1{1-\alpha\frac{k}{N-1}}+\frac \alpha N\frac{1-\frac k{N-1}}{\left(1-\alpha \frac k{N-1}\right)\left(1-\alpha \frac kN\right)}\right)\\
=1+\alpha xS_{N-1}+\frac {\alpha^2x} N\sum\limits_{k = 0}^{N-1} \binom{N-1}{k}x^{k} (1-x)^{N-k-1} \frac{1-\frac k{N-1}}{\left(1-\alpha \frac k{N-1}\right)\left(1-\alpha \frac kN\right)}
$$
This is almost a good way to compare $S_N$ with $S_{N-1}$.
