Prove a combinatorial identity: $ \sum_{n_1+\dots+n_m=n} \prod_{i=1}^m \frac{1}{n_i}\binom{2n_i}{n_i-1}=\frac{m}{n}\binom{2n}{n-m}$

Using the generating function of the Catalan numbers the left is $$[z^n] \left(-1 + \frac{1-\sqrt{1-4z}}{2z}\right)^m.$$

This is $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \left(-1 + \frac{1-\sqrt{1-4z}}{2z}\right)^m \; dz.$$

Using Lagrange inversion put $1-4z=w^2$ so that $1/4-z=1/4 \times w^2$ or $z=1/4\times(1-w^2)$ and $dz = -1/2 \times w\; dw.$ This gives for the integral $$\frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{4^{n+1}}{(1-w^2)^{n+1}} \left(-1 +\frac{1-w}{1/2\times(1-w^2)}\right)^m \times\left(-\frac{1}{2} w\right) \; dw \\ = \frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{4^{n+1}}{(1-w^2)^{n+1}} \left(\frac{w^2-1+2-2w}{1-w^2}\right)^m \times\left(-\frac{1}{2} w\right) \; dw \\ = \frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{4^{n+1}}{(1-w^2)^{n+1}} \frac{(w-1)^{2m}}{(1-w^2)^{m}} \times\left(-\frac{1}{2} w\right) \; dw \\ = \frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{4^{n+1}}{(1-w^2)^{n+m+1}} (1-w)^{2m} \times\left(-\frac{1}{2} w\right) \; dw \\ = - \frac{2^{2n+1}}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(1-w)^{n+m+1}(1+w)^{n+m+1}} (1-w)^{2m} \times w\; dw \\ = - \frac{2^{2n+1}}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(1-w)^{n-m+1}(1+w)^{n+m+1}} \times w\; dw \\ = \frac{(-1)^{n-m} 2^{2n+1}}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n-m+1}(1+w)^{n+m+1}} \times w\; dw.$$

The integral has two pieces, the first is $$\frac{(-1)^{n-m} 2^{2n+1}}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n-m}(2+w-1)^{n+m+1}} \; dw \\ = \frac{(-1)^{n-m} 2^{n-m}}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n-m}(1+(w-1)/2)^{n+m+1}} \; dw.$$

This is $$(-1)^{n-m} 2^{n-m} (-1)^{n-m-1} {n-m-1+n+m\choose n+m}\frac{1}{2^{n-m-1}} = -2{2n-1\choose n+m}.$$

The second piece is $$\frac{(-1)^{n-m} 2^{2n+1}}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n-m+1}(2+w-1)^{n+m+1}} \; dw \\ = \frac{(-1)^{n-m} 2^{n-m}}{2\pi i} \int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n-m+1}(1+(w-1)/2)^{n+m+1}} \; dw.$$

This is $$(-1)^{n-m} 2^{n-m} (-1)^{n-m} {n-m+n+m\choose n+m}\frac{1}{2^{n-m}} = {2n\choose n+m}.$$

Collecting the two pieces we obtain $${2n\choose n+m} -2{2n-1\choose n+m} = {2n\choose n+m} - 2\frac{n-m}{2n}{2n\choose n+m} \\= \frac{2n-2n+2m}{2n}{2n\choose n-m} = \frac{m}{n}{2n\choose n-m}.$$

Since $$ \sum_{n\geq 1}\frac{1}{n}\binom{2n}{n-1}x^n = -1+\frac{1-\sqrt{1-4x}}{2x}=\frac{1-\sqrt{1-4x}}{1+\sqrt{1+4x}}\tag{1}$$ the LHS is just the coefficient of $x^n$ in the Taylor series of $$ f(x) = \left(\frac{1-\sqrt{1-4x}}{1+\sqrt{1+4x}}\right)^m \tag{2}$$ that can be computed by using Chebyshev polynomials or the Lagrange inversion theorem, since the inverse function of $\frac{1-\sqrt{1-4x}}{1+\sqrt{1+4x}}$ is just $\frac{y(1-y^2)}{(1+y^2)^2}$.

We will consider the "circle of pairs" ${CP}_k$, more precisely, ${CP}_k:=\mathbb{Z}_k\times\{1,2\}$, where $k$ is some positive integer and $\mathbb{Z}_k$ is the additive cyclic group as usual.

A colouring of ${CP}_k$ is a function $f:{CP}_k\rightarrow\{0,1\}$. The size of a colouring $f$ is defined as $|f^{-1}(1)|$. Two colourings $f,g$ of ${CP}_k$ (of the same size) can be the same "by rotation", more precisely, there exists $a\in\mathbb{Z}_k$ such that $f((x+a,i))=g((x,i))$ for all $\{x,i\}\in{CP}_k$.

Note that the left hand side of the identity is exactly the number of ways to have colouring of $m$ linearly ordered circle of pairs ${CP}_{n_1},\dots,{CP}_{n_m}$ of sizes $n_1-1,\dots,n_m-1$ respectively for $n_1,\dots,n_m>0$ with $n_1+\dots+n_m=n$ up to rotation.

Before we show the right hand side of the identity is also this number, we take a more detailed look into a colouring. Let $f$ be a colouring of ${CP}_k$ of size $l$ for some $l<k$. For $x\in\mathbb{Z}_k$, we denote the pair $\{(x,1),(x,2)\}$ by $P_x$. Note that for each $x\in\mathbb{Z}_k$, $f(P_x)$ can be $\{0,1\}$, $\{0\}$ or $\{1\}$.

We now partition the pairs into $4$ types:

(i) For each $x\in\mathbb{Z}_k$, $P_x$ is of Type I if $f(P_x)=\{0,1\}$;

(ii) For each $x\in\mathbb{Z}_k$, $P_x$ is of Type II if $f(P_x)=\{1\}$;

(iii) Pairs Type III are defined in a way that each of them is one-to-one corresponded with one pair of Type II: Firstly, we mark all pairs of Type I and II. Then, for each pair $P_y$ of Type II, find the smallest $a\in\{1,\dots,k\}$ with that $P_{x+a}$ is still not marked yet, this pair $P_{x+a}$ must satisfy $f(P_{x+a})=\{0\}$, mark $P_{x+a}$;

(iv) For $x\in\mathbb{Z}_k$, $P_x$ is of Type IV if it is neither of Type I, II nor III.

The main observations are, given a colouring of ${CP}_k$ of size $l<k$, (a) although in (iii) we can pick pairs of Type II in several different orders, but it does not alter which type a pair will be assigned; (b) the total number of pairs of Types I, II and III is $l$, and the number of pairs of Type IV is $k-l$. For example, with given a colouring of ${CP}_{n_1}$ of size $n_1-1$, there is exactly one pair of Type IV and it is uniquely determined.

Back to our problem, we take a colouring $f$ of ${CP}_n$ of size $n-m$ (up to rotation). In this colouring we have exactly $m$ pairs $P_{x_1},\dots,P_{x_m}$ of Type IV. We pick one pair $P_{x_j}$ amongst these $m$ pairs. You may notice that we can have $\frac m n \binom{2n}{n-m}$ ways to do it. We now cut ${CP}_n$ into $m$ segments, namely $\{P_{x_1},\dots,P_{x_2-1}\},\{P_{x_2},\dots,P_{x_3-1}\},\dots,\{P_{x_m},\dots,P_{x_1-1}\}$, and glue them into $m$ circle of pairs in a natural way. We order the circle of pairs containing $P_{x_j}$ as the first circle of pairs, the circle of pairs containing $P_{x_{j+1}}$ as the second circle of pairs, and so on. And please excuse me for leaving the tedious detail of showing that this process does provide us a one-to-one correspondence :)