Prove $\sum_{k=j}^{\lfloor n/2\rfloor}\frac1{4^k}\binom{n}{2k}\binom{k}{j}\binom{2k}{k}=\frac1{2^n}\binom{2n-2j}{n-j}\binom{n-j}j$

Question

let $x>1$, $n\in \mathbb N$ and $$P_{n}(x)=\dfrac{1}{\pi}\int_{0}^{\pi}(x+\sqrt{x^2-1}\cos{t})^ndt.$$ Prove that $$P_{n}(x)=\dfrac{1}{\pi}\int_{0}^{\pi}\dfrac{1}{\left(x-\sqrt{x^2-1}\cos{t}\right)^{n+1}}dt.$$

In this answer, @robjohn have proved that $$P_n(x)=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{n-2k}(x^2-1)^k\frac1{4^k}\binom{2k}{k},\tag{1}$$ and $$\sum_{n=0}^\infty I_n(x)v^n=\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{v(2x-v)}{4}\right)^k,\tag{2}$$ where $$I_n(x)=\dfrac{1}{\pi}\int_{0}^{\pi}\dfrac{1}{\left(x-\sqrt{x^2-1}\cos{t}\right)^{n+1}}dt.$$

My question: How to deduce $P_n(x)=I_n(x)$ from $(1)$ and $(2)$?

My attempt: It follows from $(1)$ that \begin{align*} P_n(x)&=\sum_{k=0}^{\lfloor n/2\rfloor}\sum_{j=0}^k\binom{n}{2k}x^{n-2k}\binom{k}{j}x^{2(k-j)}(-1)^j\frac1{4^k}\binom{2k}{k}\\&=\sum_{j=0}^{\lfloor n/2\rfloor}\sum_{k=j}^{\lfloor n/2\rfloor}\binom{n}{2k}\binom{k}{j}(-1)^j\frac1{4^k}\binom{2k}{k}x^{n-2j}.\tag{3} \end{align*} It follows from $(2)$ that \begin{align*} \sum_{n=0}^\infty I_n(x)v^n=\sum_{k=0}^\infty\binom{2k}{k}\frac1{4^k}v^k(2x-v)^k=\sum_{k=0}^\infty\sum_{j=0}^k\binom{2k}{k}\frac1{4^k}\binom kj(2x)^{k-j}(-1)^jv^{k+j}. \end{align*} For any $n\in\mathbb N_{\geq0}$, if $k+j=n$ and $k\geq j$, then $n\geq 2j$ and thus $i\leq\lfloor n/2\rfloor$. Hence $$\sum_{n=0}^\infty I_n(x)v^n=\sum_{n=0}^\infty\sum_{j=0}^{\lfloor n/2\rfloor}\binom{2n-2j}{n-j}\frac1{4^{n-j}}\binom{n-j}j2^{n-2j}x^{n-2j}(-1)^jv^n,$$ and then $$I_n(x)=\sum_{j=0}^{\lfloor n/2\rfloor}\binom{2n-2j}{n-j}\frac1{2^n}\binom{n-j}j(-1)^jx^{n-2j}.\tag{4}$$ Comparing $(3)$ and $(4)$, it suffices to show that $$\sum_{k=j}^{\lfloor n/2\rfloor}\frac1{4^k}\binom{n}{2k}\binom{k}{j}\binom{2k}{k}=\frac1{2^n}\binom{2n-2j}{n-j}\binom{n-j}j,\qquad 0\leq j\leq\lfloor n/2\rfloor.\tag{5}$$ But I don't know how to prove $(5)$. Any help would be appreciated!

Answer 1

We seek to prove that

$$\sum_{k=j}^{\lfloor n/2\rfloor} \frac{1}{4^k} {n\choose 2k} {k\choose j} {2k\choose k} = \frac{1}{2^n} {2n-2j\choose n-j} {n-j\choose j}.$$

Observe that

$${n\choose 2k} {2k\choose k} = \frac{n!}{(n-2k)! \times k! \times k!} = {n\choose k} {n-k\choose n-2k}$$

so we get for the LHS

$$\sum_{k=j}^{\lfloor n/2\rfloor} {n\choose k} \frac{1}{4^k} {k\choose j} {n-k\choose n-2k}.$$

The coefficient extractor for the rightmost binomial coefficient enforces the upper range:

$$[z^n] (1+z)^n \sum_{k\ge j} {n\choose k} \frac{z^{2k}}{(1+z)^k} \frac{1}{4^k} {k\choose j} \\ = [z^n] (1+z)^n [w^j] \sum_{k\ge j} {n\choose k} \frac{z^{2k}}{(1+z)^k} \frac{1}{4^k} (1+w)^k.$$

Now when $k\lt j$ we get zero from the coefficient extractor in $w$ so it enforces the lower range:

$$[z^n] (1+z)^n [w^j] \sum_{k\ge 0} {n\choose k} \frac{z^{2k}}{(1+z)^k} \frac{1}{4^k} (1+w)^k \\ = [z^n] (1+z)^n [w^j] \left[ 1 + \frac{z^2 (1+w)}{4 (1+z)} \right]^n \\ = \frac{1}{2^{2n}} [z^n] [w^j] [4+4z+z^2(1+w)]^n \\ = \frac{1}{2^{2n}} [z^n] [w^j] [(z+2)^2 + z^2 w]^n \\ = \frac{1}{2^{2n}} [z^n] {n\choose j} z^{2j} (z+2)^{2n-2j} = \frac{1}{2^{2n}} [z^{n-2j}] {n\choose j} (z+2)^{2n-2j} \\ = \frac{1}{2^{2n}} {n\choose j} {2n-2j\choose n-2j} 2^n = \frac{1}{2^n} {n\choose j} {2n-2j\choose n-2j}.$$

Use

$${n\choose j} {2n-2j\choose n-2j} = \frac{(2n-2j)!}{(n-j)! \times j! \times (n-2j)!} = {2n-2j\choose n-j} {n-j\choose j}$$

to conclude. This proof is an instance of Egorychev method I where formal power series are sufficient.