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

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

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 $m\geq m_0$.
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 $\exp(-c)$.