How to simplify a partial sum obtained by Legendre polynomials Context
I am working on an electrostatics problem. I have undertaken Fourier analysis. By $k$, I denote a natural number $k=0,1,2,\ldots$. I have obtained the following partial sum in terms of the even number $2k$.
$$S{(2k)}=\frac1{2^{2k}}\sum_{m=0}^{k}(-1)^m\binom{2k}{m}\binom{4k-2m}{2k}\,\frac{1}{2k-2m+1}.$$
The boundary problem that this question originates from has odd symmetry. Therefore, I expect all even Fourier coefficients--except potentially $2k=0$--to be null. This expectation is verified by the correct answer below.
Question
Can the above expression be simplified further?
 A: We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ of a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n\tag{1}
\end{align*}

We obtain for $k>0$
\begin{align*}
\color{blue}{S(2k)}&=\color{blue}{\frac{1}{2^{2k}}\sum_{m=0}^k(-1)^m\binom{2k}{m}\binom{4k-2m}{2k}\frac{1}{2k-2m+1}}\\
&=\frac{1}{k2^{k+1}}\sum_{m\geq 0}(-1)^m\binom{2k}{m}\binom{4k-2m}{2k-2m+1}\tag{2}\\
&=\frac{1}{k2^{k+1}}\sum_{m\geq 0}(-1)^m\binom{2k}{m}[z^{2k-2m+1}](1+z)^{4k-2m}\tag{3}\\
&=\frac{1}{k2^{k+1}}[z^{2k+1}](1+z)^{4k}\sum_{m\geq 0}(-1)^m\binom{2k}{m}\left(\frac{z}{1+z}\right)^{2m}\tag{4}\\
&=\frac{1}{k2^{k+1}}[z^{2k+1}](1+z)^{4k}\left(1-\left(\frac{z}{1+z}\right)^2\right)^{2k}\tag{5}\\
&=\frac{1}{k2^{k+1}}[z^{2k+1}](1+2z)^{2k}\tag{6}\\
&\,\,\color{blue}{=0}\tag{7}
\end{align*}

Comment:

*

*In (2) we use the binomial identity $\binom{4k-2m}{2k}\frac{1}{2k-2m+1}=\binom{4k-2m}{2k-2m}\frac{1}{2k-2m+1}
=\binom{4k-2m}{2k-2m-1}\frac{1}{2k}$.


*In (3) we use the coefficient of operator according to (1). We also set the upper limit of the sum to $\infty$ which doesn't change the sum since we are adding zeros only.


*In (4) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.


*In (5) we apply the binomial theorem.


*In (6) we do some simplifications.


*In (7) we select the coefficient of $z^{2k+1}$.
A: We can rewrite the sum as
$$
\eqalign{
  & S(k) = {1 \over {2^{\,2k} }}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m}
 \left( \matrix{  2k \cr   m \cr}  \right)\left( \matrix{  4k - 2m \cr   2k \cr}  \right)
{1 \over {2k - 2m + 1}}}  =   \cr 
  &  = {1 \over {2^{\,2k} }}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m}
 \left( \matrix{  2k \cr   m \cr}  \right){{\left( {4k - 2m} \right)^{\,\underline {\,2k\,} } }
 \over {\left( {2k} \right)!}}{1 \over {\left( {2k - 2m + 1} \right)}}}  =   \cr 
  &  = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m}
 \left( \matrix{  2k \cr   m \cr}  \right)
\left( {2\left( {2k - m} \right)} \right)^{\,\underline {\,2k - 1\,} } }  \cr} 
$$
Now the Falling factorial
$$
p(m,k) = \left( {2\left( {2k - m} \right)} \right)^{\,\underline {\,2k - 1\,} } 
$$
is a polynomial in $m$ with the following characteristics.
$$
\left\{ {\matrix{
   {k = 0\quad  \Rightarrow \quad } & \matrix{
  p(m,0) = \left( { - 2m} \right)^{\,\underline {\, - 1\,} }
  = {1 \over {\left( { - 2m + 1} \right)^{\underline {\,1\,} } }}
 = {1 \over { - 2m + 1}}\quad  \Rightarrow  \hfill \cr 
   \Rightarrow \quad p(0,0) = 1 \hfill \cr}   \cr 
   {1 \le k\quad  \Rightarrow \quad } & \matrix{
  p(m,k) = \left( {2\left( {2k - m} \right)} \right)^{\,\underline {\,2k - 1\,} }  \hfill \cr 
  {\rm polynomial}\,{\rm in}\,{\rm m}\,{\rm ofdegree}\;2k - 1 \hfill \cr 
  {\rm with}\,{\rm zeros} \in \left[ {k + 1,\;2k} \right] \hfill \cr}   \cr 
 } } \right.
$$
Therefore we have
$$
S(0) = 1
$$
and for $1 \le k$
$$
S(k)\quad \left| {\;1 \le k} \right. = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^k
 {\left( { - 1} \right)^{\,m} \left( \matrix{  2k \cr  m \cr}  \right)p(m,k)} 
$$
Since $p(m,k) = 0$ for $k+1 \le m \le 2m$ we can extend the sum to $2k$ and multiply the summand by $1 = (-1)^{2k}$
$$
\eqalign{
  & S(k)\quad \left| {\;1 \le k} \right.\quad
  = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^{2k}
 {\left( { - 1} \right)^{\,m} \left( \matrix{  2k \cr   m \cr}  \right)p(m,k)}  =   \cr 
  &  = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^{2k}
 {\left( { - 1} \right)^{2k - \,m} \left( \matrix{  2k \cr   m \cr}  \right)p(m,k)}  \cr} 
$$
Here we recognize that the sum represents the finite difference (unitary step)
of order $2k$ of $p(m,k)$, and it is known (re. to the Newton series)
that the difference of a polynomial , of order greater than the degree of the same is identically null.
In conclusion $$S(k)= \delta _{\,k, \, 0} =\binom {0}{k}$$
A: Since (a known formula which easily follows from e.g. Rodrigues') $$P_n(x)=\frac{1}{2^n}\sum_{m=0}^{\lfloor n/2\rfloor}(-1)^m\binom{n}{m}\binom{2n-2m}{n}x^{n-2m},$$ the given expression is equal to $\int_0^1 P_{2k}(x)\,dx=\frac12\int_{-1}^1 P_{2k}(x)\,dx$.
This is $1$ for $k=0$ and $0$ for $k>0$.
