If we did not know it before, then wikipedia teaches us the [generalized geometric series](https://en.wikipedia.org/wiki/Binomial_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)?



Optimal scaling of Lipschitz estimates in generalized geometric series
$\newcommand\ep\varepsilon\newcommand\de\delta$
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](https://en.wikipedia.org/wiki/Negative_binomial_distribution#:%7E:text=In%20probability%20theory%20and%20statistics,failures%20(denoted%20r)%20occurs.) 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.


