<s>[INST] Prove that $\sum_{k=0}^{2n} \binom {2n+k}{k} \binom{2n}{k} \frac{(-1)^k}{2^k} \frac{1}{k+1} = 0. $ 
Let $n$ be a positive integer. Prove that
  $$
\sum_{k=0}^{2n} \binom {2n+k}{k} \binom{2n}{k} \frac{(-1)^k}{2^k} \frac{1}{k+1} = 0.
$$

I am trying to solve this by using induction on $n$. I have proven the sum to be zero in the case $n=1$. Assuming that the sum is zero for $n=m$ ($m$ is a positive integer), how do I prove that it implies that the sum is zero for $n=m+1$?
Can I get some hints?
 [/INST] Let's approach the sum through the Hypergeometric Function.
To this purpose let's rewrite it as
$$
\eqalign{
  & S(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,2n} \right)} {\binom{2n+k}{k} \binom{2n}{k} 
{{\left( { - 1} \right)^{\,k} } \over {2^{\,k} \left( {k + 1} \right)}}}  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,2n} \right)} {\binom{2n+k}{k} \binom{2n}{k} 
 {1 \over {\left( {k + 1} \right)}}\left( { - {1 \over 2}} \right)^{\,k} }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,2n} \right)} {t_{\,k} \left( { - {1 \over 2}} \right)^{\,k} }  \cr} 
$$
The $t_k$ are in the following ratio
$$
\eqalign{
  & t_{\,0}  = 1  \cr 
  & {{t_{\,k + 1} } \over {t_{\,k} }} =   \cr 
  &  = {{\left( {2n + k + 1} \right)!} \over {\left( {k + 1} \right)!\left( {k + 1} \right)!\left( {2n - k - 1} \right)!\left( {k + 2} \right)}}
  {{k!k!\left( {2n - k} \right)!\left( {k + 1} \right)} \over {\left( {2n + k} \right)!}} =   \cr 
  &  =  - {{\left( {k + 2n + 1} \right)\left( {k - 2n} \right)} \over {\left( {k + 2} \right)}}{1 \over {\left( {k + 1} \right)}} \cr} 
$$
so the sum can be expressed as
$$
\eqalign{
 & S(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,2n} \right)} {\binom{2n+k}{k} \binom{2n}{k} 
{{\left( { - 1} \right)^{\,k} } \over {2^{\,k} \left( {k + 1} \right)}}}  =   \cr 
  &  = {}_2F_{\,1} \left( {\left. {\matrix{   {2n + 1,\; - 2n}  \cr    2  \cr  } \,} \right|1/2} \right) \cr} 
$$
For $n=0$ this gives
$$
S(0) = {}_2F_{\,1} \left( {\left. {\matrix{   {1,\;0}  \cr    2  \cr  } \,} \right|1/2} \right) = 1
$$
while for $0<n$ we have
$$
\eqalign{
  & {}_2F_{\,1} \left( {\left. {\matrix{    {2n + 1,\; - 2n}  \cr     2  \cr   } \,} \right|1/2} \right)\quad \left| {\;0 < n} \right.\quad  =   \cr 
  &  = {{\Gamma \left( 2 \right)} \over {\Gamma \left( {2n + 1} \right)\Gamma \left( { - 2n} \right)}}\sum\limits_{0\, \le \,k\,}
 {{{\Gamma \left( {2n + 1 + k} \right)\Gamma \left( { - 2n + k} \right)} \over {\Gamma \left( {2 + k} \right)}}} {1 \over {2^{\,k} k!}} \cr} 
$$
Note that we can arrive to the same result by expressing the binomials through the Gamma function and
performing some algebraic simplifications.
To the fraction outside the sum we can apply the Reflection formula for the Gamma function,
which in the inverted form is valid all over the complex field
$$
{1 \over {\Gamma \left( {z + 1} \right)\,\Gamma \left( { - z} \right)}} =  - {{\sin \left( {\pi \,z} \right)} \over \pi }\quad \left| {\;\forall z \in \mathbb C} \right.
$$
then clearly
$$
S(n)\quad \left| {\;0 < n \in Z} \right. = \sin \left( {2\pi \,n} \right) \cdot \left(  \cdots  \right) = 0
$$
</s>