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I need to find a good, computable upper bound for the expression $$\sum_{k=0}^m x^k \binom{n+k}{k}$$ as a function of $x$, $n$ and $m$, and where $0<x< 1/2$ is real, and $0<n\leq m$ are integers. I would love to hear some ideas.

Remark A. I know that $\sum_{k=0}^\infty x^k \binom{n+k}{k}=(1-x)^{-n-1}$. So my question is how to bound the first $m$ terms of this infinite sum.

Remark B. What I'm actually trying to do is bound the sum $\sum_{k=1}^m x^k C_{m-n+1}(n,k)$, where $C_{d}(n,k)$ is a Catalan trapezoid number of order $d$. I am then using the bound $C_d(n,k)\leq \binom{n+k}{k}$, which is tight for low values of $k$ but loose for high values. If someone knows other bounds for Catalan trapezoid numbers, that could be useful here, let me know. For instance, it could be useful to have results about generating functions related to Catalan trapezoid numbers.

Alfonso Cevallos
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  • For large values of $n$, you could try the combo of Stirling's approximation along with the Saddle point lemma. Essentially rewrite the sum as an integral and use $$\int e^{nf(x)}dx \sim e^{nf(z)}$$ where $z$ is where the function attains it's maximum on the interval over which you are integrating – Kesav Krishnan Jul 16 '21 at 14:35
  • You can use the bound $\binom{m}{r}\le (em/r)^r$ (see first line of Wikipedia article on bounds for binomial coefficients) as follows: $$ \sum_{k=0}^m x^k\binom{n+k}{k}\le \sum_{k=0}^nx^k(e(n+k)/k)^k=\sum_{k=0}^m(ex)^k\cdot \left(1+\frac{n}k\right)^k\le \sum_{k=0}^m(ex)^k\cdot e^n=e^n\frac{(ex)^{m+1}-1}{ex-1}. $$ This works as long as $x\neq 1/e$. When $x=1/e$, the bound is instead $(m+1)e^n$. – Mike Earnest Jul 16 '21 at 15:42

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