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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 $$f(m) \leq g(m) ?$$

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    $\begingroup$ $\sum_{n = 1}^{\infty}{n + m\,\, \choose n}\,\,\,\mathrm{e}^{-cn} = \sum_{n = 1}^{\infty}{-n - m + n - 1\,\,\,\,\, \choose n\,\,}\,\,\,\left(-1\right)^{n}\left(\,\mathrm{e}^{-c}\,\right)^{n} = \sum_{n = 1}^{\infty}{-m - 1\,\,\, \choose n}\,\,\,\left(-\,\mathrm{e}^{-c}\,\right)^{n} = \left(1 - \,\mathrm{e}^{-c}\,\,\right)^{-m - 1}$. $\endgroup$ – Felix Marin Jul 27 '16 at 20:15
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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 $m\geq m_0$.

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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 $\exp(-c)$.

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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\,\mathcal{#1}} \newcommand{\mrm}[1]{\,\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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