Optimal scaling of Lipschitz estimates in generalized geometric series If we did not know it before, then wikipedia teaches us the generalized geometric series
$$\sum_{n \ge 0} \binom{n+k}{n} (1-\mu)^{n} \mu^k = \frac{1}{\mu}.$$
We can then study for $0 <\varepsilon < \mu,\nu  <1-\varepsilon$ the Lipschitz bound
$$\sum_{n \ge 0} \binom{n+k}{n} \vert (1-\mu)^{n} \mu^k - (1-\nu)^n \nu^k \vert \le C_{\varepsilon}(k)\vert \mu-\nu\vert,$$
and I wonder how this bound, or more precisely the constant $C_{\varepsilon}(k)$ will depend on $k$? (is it constant, polynomially, exponentially)?
 A: $\newcommand\ep\varepsilon\newcommand\de\delta$
The best possible value for the Lipschitz constant $C_\ep(k)$ is
\begin{equation*}
    C^*_\ep(k):=\sup_{\ep<\mu<1-\ep}c_\mu(k),\tag{-1}
\end{equation*}
where
\begin{align*}
    c_\mu(k)&:=\sum_{n\ge0} \binom{n+k}n \Big|\frac d{d\mu}{(1-\mu)^n} \mu^k\Big| \\ 
    &=\sum_{n\ge0} \binom{n+k}{k} \mu ^{k-1}(1-\mu )^{n-1} |\mu n-(1-\mu)k| \\
    &=\frac1{(1-\mu)\mu^2} E|Y_k|, \tag{0}
\end{align*}
\begin{equation*}
    Y_k:=\mu X_{k+1}-(1-\mu)k,
\end{equation*}
and $X_l$ is a random variable with the negative binomial distribution with parameters $l,1-\mu$.
Further,
\begin{equation*}
    E|Y_k|\le\sqrt{EY_k^2}=\sqrt{1-\mu}\sqrt{k+2-\mu}\le\sqrt{1-\mu}\sqrt{k+2}. \tag{1}
\end{equation*}
So,
\begin{equation*}
    C^*_\ep(k)\le\frac{\sqrt{k+2}}{\ep^2\sqrt{1-\ep}}. 
\end{equation*}
Working a bit harder, we can find the asymptotics for $c_\mu(k)$ and hence for $C^*_\ep(k)$. Indeed, since $X_{k+1}$ is equal in distribution to the sum of $k+1$ iid copies of $X_1$, we can use the central limit theorem to see that
\begin{equation*}
    Z_k:=\frac{Y_k-(1-\mu)}{\mu\sqrt{k+1}}
    =\sqrt{k+1}\,\Big(\frac{X_{k+1}}{k+1}-\frac{1-\mu}\mu\Big) \tag{2}
    \to Z
\end{equation*}
in distribution (as $k\to\infty$), where $Z\sim N(0,\frac{1-\mu}{\mu^2})$.
Also, $EZ_k^2=O(1)$ (cf. (1)), so that the $Z_k$'s are uniformly integrable. So, by (2),
\begin{equation}
    E|Z_k|\to E|Z|=\sqrt{\frac{1-\mu}{\mu^2}}\sqrt{\frac2\pi},
\end{equation}
which implies
\begin{equation}
    E|Y_k|\sim\mu\sqrt k\,\sqrt{\frac{1-\mu}{\mu^2}}\sqrt{\frac2\pi},
\end{equation}
whence, by (0),
\begin{equation}
    c_\mu(k)\sim\sqrt k\,\sqrt{\frac2\pi}\frac1{\mu^2\sqrt{1-\mu}}
\end{equation}
and, by (-1),
\begin{equation*}
    C^*_\ep(k)\sim\sqrt k\,\sqrt{\frac2\pi}\frac1{\ep^2\sqrt{1-\ep}}, 
\end{equation*}
which is in agreement with my other answer on this page.
A: Here is a partial answer:
$$C_{\varepsilon}(k)\ge
\sum_{n \ge 0} \binom{n+k}{n} \lim_{\mu\to1/2,\nu\to1/2}
\Big|\frac{(1-\mu)^n \mu^k - (1-\nu)^n \nu^k}{\mu-\nu}\Big| \\ 
=\sum_{n \ge 0} \binom{n+k}{n} \Big|
\frac d{d\mu}{(1-\mu)^n} \mu^k\Big|\,\bigg|_{\mu=1/2} \\ 
=:s_2(k)=-2^{1-2 k} \left(2^{2 k+1}-\binom{2 k+1}{k} \, _2F_1\left(2,2 k+2;k+2;\frac{1}{2}\right)\right) 
$$
and $s_2(k)\sim8\sqrt k/\sqrt\pi$ as $k\to\infty$.
The above expression for $s_2(k)$ and its asymptotics were obtained with Mathematica.
It now seems plausible that $C_{\varepsilon}(k)$ grows as $\sqrt k$.
