help computing or simplifying $\sum_{1}^{n}\binom{n}{k}\binom{n}{k-1}k$

Question

for $\beta > 4$, prove that: $\displaystyle\lim_{n\rightarrow \infty} \frac{\sum_{1}^{n}\binom{n}{k}\binom{n}{k-1}k}{\beta^{n}}$

so far I got that this expression is equal to: $n\sum_{1}^{n}\binom{n-1}{k-1}\binom{n}{k-1}$ but I have no clue for how to continue the complete proof

Answer 1

Hint: use Vandermonde's thm: $$n\sum _{k=0}^{n-1}\binom{n-1}{k}\binom{n}{n-k}=n\binom{2n-1}{n}=\frac{2n}{2}\binom{2n-1}{n}=\frac{n}{2}\binom{2n}{n},$$ Use that $\binom{2n}{n}\sim \frac{4^n}{\sqrt{n\cdot \pi}}.$

Answer 2

For $k\ge1,$

$$k\binom nk=\cdots=n\binom{n-1}{k-1}$$

Now compare the coefficients of $x^{n-1}$ in $$(1+x)^{n-1}(x+1)^{n-1}=(1+x)^{2n-2}$$

$$\sum_{k=1}^n(\binom{n-1}{k-1})^2=\binom{2n-2}{n-1}$$

Now use https://en.m.wikipedia.org/wiki/Stirling%27s_approximation

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 1}^{n}{n \choose k}{n \choose k - 1}k & = \sum_{k = 1}^{n}{n \choose k}k{n \choose n - k + 1} = \sum_{k = 1}^{n}{n \choose k}k\bracks{z^{n - k +1}}\pars{1 + z}^{n} \\[5mm] & = \bracks{z^{n}}\pars{1 + z}^{n}\sum_{k = 1}^{n}{n \choose k}kz^{k - 1} = \bracks{z^{n}}\pars{1 + z}^{n}\,\partiald{}{z} \sum_{k = 1}^{n}{n \choose k}z^{k} \\[5mm] & = \bracks{z^{n}}\pars{1 + z}^{n}\,\partiald{\pars{1 + z}^{n}}{z} = \bracks{z^{n}}\pars{1 + z}^{n}\, n\pars{1 + z}^{n - 1} \\[5mm] & = n\bracks{z^{n}}\pars{1 + z}^{2n - 1} = \bbx{n{2n - 1 \choose n}} = {\pars{2n - 1}! \over \bracks{\pars{n - 1}!}^{\, 2}} \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {\root{2\pi}\pars{2n - 1}^{2n - 1/2}\expo{-2n + 1} \over \bracks{\root{2\pi}\pars{n - 1}^{n - 1/2}\expo{-n + 1}}^{\, 2}} \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {1 \over \root{2\pi}}\,{2^{2n - 1/2}n^{2n - 1/2}\, \bracks{1 - 1/\pars{2n}}^{2n} \over \bracks{n^{n - 1/2}\pars{1 - 1/n}^{n}}^{\, 2}}\,\expo{-1} \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {2^{2n - 1} \over \root{\pi}}\,{n^{1/2}\, \expo{-1} \over \pars{\expo{-1}}^{\, 2}}\,\expo{-1} = {1 \over \root{\pi}}\,2^{2n - 1}\, n^{1/2} \end{align} The coveted limit becomes $$ \left.\lim_{n \to \infty}\pars{{1 \over \root{\pi}}\,2^{2n - 1}\, n^{1/2}}/ \beta^{n}\right\vert_{\ \beta\ >\ 4} = {1 \over 2\root{\pi}}\lim_{n \to \infty}\pars{4 \over \beta}^{n}n^{1/2} = \bbx{\large\color{red}{0}} $$