Upper bound to a series with binomial coefficients Let $c>0$ and $m$ be a positive integer. The following sum is convergent, but how fast does it grow with $m$ as $m$ is large? 
$$
f(m)= \sum\limits_{n=1}^{\infty} \binom{n + m}{n} e^{-c \,  n}
$$
Is there a polinom in $m$, $g(m)$, such that
$$f(m) \leq g(m) ?$$
 A: Put $$f_m(x)=\sum_{n\geq 1}\binom{n+m}{n}x^n=\frac{1}{m!}\sum_{n\geq 1}(n+m)(n+m-1)\cdots (n+1)x^n=\frac{g_m(x)}{m!}$$
We have:
$$g_m(x)=(\frac{d}{dx})^m(\sum_{n\geq 1} x^{n+m})=(\frac{d}{dx})^m(\frac{x^{m+1}}{1-x})$$
As $$\frac{x^{m+1}}{1-x}=-(1+x+\cdots+x^m)+\frac{1}{1-x}$$, this gives that $g_m(x)=-m!+\frac{m!}{(1-x)^{m+1}}$. It is easy to finish, replacing $x$ by $\exp(-c)$. 
A: By stars and bars for any $x\in(0,1)$ and $m\in\mathbb{N}^*$ we have:
$$ \frac{1}{(1-x)^{m+1}}=\sum_{n\geq 0}\binom{m+n}{n}x^n \tag{1}$$
hence:
$$ \sum_{n\geq 1}\binom{m+n}{n}e^{-cn} = \color{red}{\frac{1}{(1-e^{-c})^{m+1}}-1}.\tag{2} $$
As a function of $m$, the RHS of $(2)$ has an exponential behaviour, hence there is no polynomial $g(m)$ that is an upper bound for the RHS of $(2)$ for any $m\geq m_0$.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\mrm{f}\pars{m}} & \equiv
\sum_{n = 1}^{\infty}{n + m \choose n}\expo{-c\, n}
\\[5mm] & =
\sum_{n = 1}^{\infty}{-\pars{n + m} + n - 1 \choose n}\pars{-1}^{n}\expo{-c\, n}
\quad\pars{~Binomial "Negation"~}
\\[5mm] & =
\sum_{n = 1}^{\infty}{-m - 1 \choose n}\pars{-\expo{-c}}^{n} =
\bracks{1 + \pars{-\expo{-c}}}^{-m - 1}\,\,\, -\,\,\, 1\quad
\pars{~Newton\ Binomial~}
\\[5mm] & =
\color{#f00}{{1 \over \pars{1 - \expo{-c}}^{m + 1}} - 1}
\end{align}
