¿Cómo encontrar? $\sum _{k=1}^{n+1} \frac{n \binom{n}{k-1}}{\binom{2 n}{k}}$

Question

Agradecería que alguien me ayudara con el siguiente problema:

P: ¿Cómo se encuentra? $$\sum _{k=1}^{n+1} \frac{n \binom{n}{k-1}}{\binom{2 n}{k}}$$

Mi trabajo usando Mathematica 11 encuentra $$\sum _{k=1}^{n+1} \frac{n \binom{n}{k-1}}{\binom{2 n}{k}}=\frac{2n+1}{n+1}$$

Answer 1

A partir de

$$n\sum_{k=1}^{n+1} {2n\choose k}^{-1} {n\choose k-1}$$

encontramos

$$n\sum_{k=1}^{n+1} \frac{n!}{(k-1)! \times (n+1-k)!} \frac{k! \times (2n-k)!}{(2n)!} \\ = \frac{n\times n!}{(2n)!} \sum_{k=1}^{n+1} \frac{k}{(n+1-k)!} (2n-k)! \\ = \frac{n!^2}{(2n)!} \sum_{k=1}^{n+1} k {2n-k\choose n+1-k}.$$

Esto es

$$\frac{n!^2}{(2n)!} \sum_{k=0}^{n+1} k [z^{n+1-k}] (1+z)^{2n-k} \\ = \frac{n!^2}{(2n)!} [z^{n+1}] (1+z)^{2n} \sum_{k=0}^{n+1} k z^k (1+z)^{-k}$$

Ahora, cuando $k\gt n+1$ no hay contribución al coeficiente y obtenemos

$$\frac{n!^2}{(2n)!} [z^{n+1}] (1+z)^{2n} \sum_{k\ge 0} k z^k (1+z)^{-k} \\ = \frac{n!^2}{(2n)!} [z^{n+1}] (1+z)^{2n} \frac{z/(1+z)}{(1-z/(1+z))^2} \\ = \frac{n!^2}{(2n)!} [z^{n+1}] (1+z)^{2n+2} \frac{z}{1+z} \\ = \frac{n!^2}{(2n)!} [z^{n}] (1+z)^{2n+1} = \frac{n!^2}{(2n)!} {2n+1\choose n} \\ = n!^2 \times (2n+1) \times \frac{1}{n! \times (n+1)!} = \frac{2n+1}{n+1}.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 1}^{n + 1}{n{n \choose k - 1} \over {2n \choose k}} & = n\sum_{k = 0}^{n}{{n \choose k} \over {2n \choose k + 1}} \\[5mm] & = n\sum_{k = 0}^{n}{n \choose k} {1 \over \pars{2n}!/\bracks{\pars{k + 1}!\pars{2n - k - 1}!}} \\[5mm] & = n\pars{2n + 1}\sum_{k = 0}^{n} {n \choose k}{\Gamma\pars{k + 2}\Gamma\pars{2n - k}! \over \Gamma\pars{2n + 2}} \\[5mm] & = n\pars{2n + 1}\sum_{k = 0}^{n} {n \choose k}\int_{0}^{1} t^{k + 1}\pars{1 - t}^{2n - k - 1}\,\dd t \\[5mm] & = n\pars{2n + 1}\int_{0}^{1}\pars{1 - t}^{2n}\, {t \over 1 - t}\sum_{k = 0}^{n}{n \choose k} \pars{t \over 1 - t}^{k}\,\dd t \\[5mm] & = n\pars{2n + 1}\int_{0}^{1}\pars{1 - t}^{2n - 1}\, t \pars{1 + {t \over 1 - t}}^{n}\,\dd t \\[5mm] & = n\pars{2n + 1}\int_{0}^{1}t\, \pars{1 - t}^{n - 1}\,\dd t \\[5mm] & = n\pars{2n + 1}\int_{0}^{1}\pars{1 - t}\,t^{n - 1}\,\dd t \\[5mm] & = n\pars{2n + 1}\pars{{1 \over n} - {1 \over n +1}} = \bbx{2n + 1 \over n + 1} \end{align}