help computing or simplifying $\sum_{1}^{n}\binom{n}{k}\binom{n}{k-1}k$ 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
 A: 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 $\binom{2n}{n}\sim \frac{4^n}{\sqrt{n\cdot \pi}}.$
A: 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
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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
$$
\left.\lim_{n \to \infty}\pars{{1 \over \root{\pi}}\,2^{2n - 1}\, n^{1/2}}/
\beta^{n}\right\vert_{\ \beta\ >\ 4} =
{1 \over 2\root{\pi}}\lim_{n \to \infty}\pars{4 \over \beta}^{n}n^{1/2} =
\bbx{\large\color{red}{0}}
$$
