Proof of (complicated?) summation equality This is a simplified case of something I'm trying to prove. 
Suppose that $N,h$ are even. I want to show that
$$ \sum_{k=1}^{(N-h)/2} \frac{2^{2k}}{kN^{\underline{2k}}}\left(\frac{N}{2}\right)^{\underline{k}}\left(\frac{N-h}{2}\right)^{\underline{k}} = \sum_{j=1}^{(N-h)/2} \frac{2}{2j+h-1} $$
where $n^{\underline{k}} = n(n-1)...(n-k+1)$ is the falling factorial.
Now the LHS terms can be written as follows
\begin{align} \frac{2^{2k}}{kN^{\underline{2k}}}\left(\frac{N}{2}\right)^{\underline{k}}\left(\frac{N-h}{2}\right)^{\underline{k}} &= \frac{1}{kN^{\underline{2k}}} \prod_{j=0}^{k-1} (N-2j)(N-h-2j)
\\&= \frac{1}{k} \prod_{j=0}^{k-1} \frac{N-h-2j}{N-2j-1}
\\&= \frac{1}{k} \prod_{j=0}^{k-1} \left(1-\frac{h-1}{N-2j-1}\right).
\end{align}
Note the RHS can be written as $H_{\frac{N}{2}-\frac{1}{2}}-H_{\frac{h}{2}-\frac{1}{2}}$ where $H_n$ is the Harmonic number.
 A: Consider the sum
$$
S^n_m=\sum_{k=1}^m4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}};\quad n\ge m\ge1.\tag{1}
$$
We are going to prove:
$$
S^n_m=\frac{2m}{2n-2m+1}.\tag{2}
$$
It is easy to check that for $m=1$ and arbitrary $n\ge1$ the expression (2) is valid:
$$
S^{n}_1=4^1\frac{\binom{n}{1}\binom{1}{1}}{\binom{2n}{2}\binom{2}{1}}=\frac{2}{2n-1}.
$$
Assume that expression (2) is valid for some $S^{n-1}_{m-1}$ $(n\ge m\ge 2)$. It follows then that it is valid for $S^{n}_{m}$ as well:
$$\begin {align}
S^{n}_{m}&=\sum_{k=1}^{m}4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}}\\
&=\sum_{k=1}^{m}4^k\frac{\frac{n}{k}\binom{n-1}{k-1}\frac{m}{k}\binom{m-1}{k-1}}{\frac{2n(2n-1)}{2k(2k-1)}\binom{2n-2}{2k-2}\frac{2k(2k-1)}{k^2}\binom{2k-2}{k-1}}\\
&=\frac{2m}{2n-1}\sum_{k=1}^{m}4^{k-1}\frac{\binom{n-1}{k-1}\binom{m-1}{k-1}}{\binom{2n-2}{2k-2}\binom{2k-2}{k-1}}\\
&=\frac{2m}{2n-1}\sum_{k=0}^{m-1}4^{k}\frac{\binom{n-1}{k}\binom{m-1}{k}}{\binom{2n-2}{2k}\binom{2k}{k}}\\
& =\frac{2m}{2n-1}\left(S^{n-1}_{m-1}+1\right)\\
&\stackrel{I.H.}{=}\frac{2m}{2n-1}\left(\frac{2(m-1)}{2(n-1)-2(m-1)+1}+1\right)\\
& =\frac{2m}{2n-1}\cdot\frac{2n-1}{2n-2m+1}\\
&=\frac{2m}{2n-2m+1},\end {align}
$$
where $\stackrel{I.H.}{=}$ means substitution of the induction assumption. Thus, the equality (2) is proved.
The rest is rather straightforward. Let $l$ be an integer number $(1\le l\le n)$. Then:
$$\begin {align}
\sum_{m=1}^l\frac{2}{2n-2m+1}&\stackrel {(2)}=\sum_{m=1}^l\frac{1}{m}\sum_{k=1}^m4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}}\\
&=
\sum_{m=1}^l\frac{1}{m}\sum_{k=1}^\infty4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}}\\
&=\sum_{k=1}^\infty4^k\frac{\binom{n}{k}}{\binom{2n}{2k}\binom{2k}{k}}\sum_{m=1}^l\frac{1}{m}\binom{m}{k}\\
&\stackrel{!}{=}\sum_{k=1}^\infty\frac{4^k}{k}\frac{\binom{n}{k}\binom{l}{k}}{\binom{2n}{2k}\binom{2k}{k}}\\
&=\sum_{k=1}^l\frac{4^k}{k}\frac{\binom{n}{k}\binom{l}{k}}{\binom{2n}{2k}\binom{2k}{k}}.\tag{3}\end {align}
$$
The proof of identity:
$$
\sum_{m=1}^l\frac{1}{m}\binom{m}{k}=\frac{1}{k}\binom{l}{k};\quad k\ge1,
$$
used in $\stackrel{!}{=}$, is left as an exercise.
The equality (3) is identical to that one claimed in question. To see it redefine in the latter $N=2n$, $h=2(n-l)$ and reverse the summation order in RHS: $j\mapsto l+1-j$.
A: To complement the nice answer provided by user355705, please refer to the answer provided in this related post where it is shown that
$$ \bbox[lightyellow] {  
\eqalign{
  & \sum\limits_{0\, \le \,k\, \le \,m} {4^{\,k} {{\left( \matrix{
  n \cr 
  k \cr}  \right)\left( \matrix{
  m \cr 
  k \cr}  \right)} \over {\left( \matrix{
  2n \cr 
  2k \cr}  \right)\left( \matrix{
  2k \cr 
  k \cr}  \right)}}}  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {{{\left( \matrix{
  m \cr 
  k \cr}  \right)} \over {\left( \matrix{
  n - 1/2 \cr 
  k \cr}  \right)}}}  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{
  m \cr 
  k \cr}  \right)1^{\,\overline {\,k\,} } \left( {n + 1/2} \right)^{\,\overline {\, - \,k\,} } }   \cr 
  &  = {1 \over {\left( {n - m + 1/2} \right)^{\,\overline {\,m\,} } }}\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{
  m \cr 
  k \cr}  \right)1^{\,\overline {\,k\,} } \left( {n - m + 1/2} \right)^{\,\overline {\,m - \,k\,} } }  =   \cr 
  &  = {{\left( {n - m + 3/2} \right)^{\,\overline {\,m\,} } } \over {\left( {n - m + 1/2} \right)^{\,\overline {\,m\,} } }} = {{n + 1/2} \over {n - m + 1/2}} = {{2n + 1} \over {2(n - m) + 1}} \cr} 
 } $$
This identity  is valid for any non-negative integer $m$, and for $n$ that can take even a real or complex value,
except for $n=m-1/2$.
This under the acception that the two binomials in $n$, when null get simplified. That is that
the two binomials be rewritten as above in terms of Falling Factorials (or Gamma function) with $n$ real, 
and simplify the fraction (take the limit).
