Let $n$ be a positive integer. Then, \begin{align} \sum_{k=1}^{n+1} \dfrac{n \binom{n}{k-1}}{\binom{2 n}{k}}=\frac{2n+1}{n+1} . \end{align}
Note that we can rewrite $\dfrac{n \binom{n}{k-1}}{\binom{2 n}{k}}$ as $\dfrac{n!^2 k \binom{2n-k}{n-1}}{(2n)!}$ (by the standard $\dbinom{a}{b} = \dfrac{a!}{b!\left(a-b\right)!}$ formula). Thus, the question is equivalent to proving that \begin{align} \sum_{k=1}^{n+1} k \dbinom{2n-k}{n-1} = \dbinom{2n+1}{n+1} . \end{align}
Starting from
$$n\sum_{k=1}^{n+1} {2n\choose k}^{-1} {n\choose k-1}$$
we find
$$n\sum_{k=1}^{n+1} \frac{n!}{(k-1)! \times (n+1-k)!} \frac{k! \times (2n-k)!}{(2n)!} \\ = \frac{n\times n!}{(2n)!} \sum_{k=1}^{n+1} \frac{k}{(n+1-k)!} (2n-k)! \\ = \frac{n!^2}{(2n)!} \sum_{k=1}^{n+1} k {2n-k\choose n+1-k}.$$
This is
$$\frac{n!^2}{(2n)!} \sum_{k=0}^{n+1} k [z^{n+1-k}] (1+z)^{2n-k} \\ = \frac{n!^2}{(2n)!} [z^{n+1}] (1+z)^{2n} \sum_{k=0}^{n+1} k z^k (1+z)^{-k}$$
Now when $k\gt n+1$ there is no contribution to the coefficient extractor and we get
$$\frac{n!^2}{(2n)!} [z^{n+1}] (1+z)^{2n} \sum_{k\ge 0} k z^k (1+z)^{-k} \\ = \frac{n!^2}{(2n)!} [z^{n+1}] (1+z)^{2n} \frac{z/(1+z)}{(1-z/(1+z))^2} \\ = \frac{n!^2}{(2n)!} [z^{n+1}] (1+z)^{2n+2} \frac{z}{1+z} \\ = \frac{n!^2}{(2n)!} [z^{n}] (1+z)^{2n+1} = \frac{n!^2}{(2n)!} {2n+1\choose n} \\ = n!^2 \times (2n+1) \times \frac{1}{n! \times (n+1)!} = \frac{2n+1}{n+1}.$$
The equality $ \sum_{k=1}^{n+1} k \binom{2n-k}{n-1} = \binom{2n+1}{n+1} $ is just a special case of $$ \sum_{k=i}^{m-j} \binom{k}{i}\binom{m-k}{j}=\binom{m+1}{i+j+1}. $$ where $i\gets1,j\gets(n-1),m\gets 2n$. For a combinatorial proof:
How many subsets of $\{1,2,\dots,m+1\}$ have size $i+j+1$?
How many such subsets contain $k+1$, and have exactly $i$ elements smaller than $k+1$?
See $\sum_{k=-m}^{n} \binom{m+k}{r} \binom{n-k}{s} =\binom{m+n+1}{r+s+1}$ using Counting argument.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
First, we show the more general result:
$$ \sum_{k=0}^{m}\binom{k}{l}\binom{m-k}{q} = \binom{m+1}{l+q+1}, \quad \text{for integers } m, l, q \ge 0. $$
Proof. $\quad$ Note that the sequence $(c_m)$ defined by $c_m=\sum_{k=0}^{m}\binom{k}{l}\binom{m-k}{q}$ is the convolution of $(a_k)$ and $(b_k),$ where $a_k=\binom{k}{l}$ and $b_k=\binom{k}{q}.$ Thus, $c_m$ is the coefficient of $z^m$ in $A(z)B(z),$ where $A(z)$ and $B(z)$ are the generating functions of $(a_k)$ and $(b_k),$ respectively. But, $A(z)=z^l/(1-z)^{l+1}$ and $B(z)=z^q/(1-z)^{q+1},$ so that $$ A(z)B(z) = \frac{z^{l+q}}{(1-z)^{l+q+2}}=\frac{1}{z}\frac{z^{l+q+1}}{(1-z)^{l+q+2}}= \frac{1}{z}\sum_{k\ge0}\binom{k}{l+q+1}z^k. $$ Finally, $$ c_m=[z^m]A(z)B(z)= \binom{m+1}{l+q+1}. \tag*{$\blacksquare$} $$
Setting $m:=2n, l:=1,$ and $q:= n-1,$ we get: $$ \sum_{k=0}^{2n}\binom{k}{1}\binom{2n-k}{n-1} = \binom{2n+1}{n+1}, $$ and it is easy to see that $\sum_{k=0}^{2n}\binom{k}{1}\binom{2n-k}{n-1} = \sum_{k=1}^{n+1}k\binom{2n-k}{n-1}.$
This is late, but here's an elementary solution that uses only summation by parts and the hockey stick identity.
We want to prove that \begin{equation}\label{3}\displaystyle \sum_{k=1}^{n+1}k\binom{2n-k}{n-1}= \binom{2n+1}{n+1}\end{equation}
Let \begin{align}S_{m} = \displaystyle \sum_{k=1}^{m}\binom{2n-k}{n-1} &= \displaystyle \sum_{r=n-1}^{2n-1}\binom{r}{n-1} - \displaystyle \sum_{r=n-1}^{2n-m-1}\binom{r}{n-1}\\& = \binom{2n}{n} - \binom{2n-m}{n} \end{align}
where the latter equality follows from the hockey stick identity.
Using summation by parts on $\displaystyle \sum_{k=1}^{n+1}k\binom{2n-k}{n-1}$, we get
\begin{align}\displaystyle \sum_{k=1}^{n+1}k\binom{2n-k}{n-1} &= (n+1)S_{n+1} - \displaystyle \sum_{k=1}^{n}S_{k}\\&= (n+1)\binom{2n}{n} - n\binom{2n}{n}+ \displaystyle \sum_{k=1}^{n}\binom{2n-k}{n}\\&= \binom{2n}{n}+ \displaystyle \sum_{r=n}^{2n-1}\binom{r}{n}\\&= \binom{2n}{n}+ \binom{2n}{n+1}\\&= \binom{2n+1}{n+1}\end{align}
where we used the hockey stick identity again in the second last step and Pascal's identity in the last step.
