Somewhat challenging binomial identity

Question

The binomial identity valid for positive integer values $n$: \begin{align*} \color{blue}{\frac{n}{4}\sum_{k=1}^n\frac{(-16)^k}{k(n+k)}\binom{n+k}{2k}\binom{2k}{k}^{-1}=\sum_{k=1}^n\frac{1}{1-2k}}\tag{1} \end{align*} is somewhat difficult to prove. I'm trying to show (1) using generating functions, but wasn't successful so far. So, I'm kindly asking for support.

Some information around the problem:

• We can write $\frac{n}{k(n+k)}=\frac{1}{k}-\frac{1}{n+k}$ which might help to split the left-hand side into simpler sums.

• If we set $a_n(z)=\sum_{k=0}^n \binom{n+k}{2k}\binom{2k}{k}^{-1}z^k$ we have \begin{align*} a_n(z)+a_{n-1}(z)&=\sum_{k=0}^n\frac{2n}{n+k}\binom{n+k}{2k}\binom{2k}{k}^{-1}z^k\\ &=2\sum_{k=0}^n\left(\binom{n+k}{2k}-\frac{1}{2}\binom{n+k-1}{2k-1}\right)\binom{2k}{k}^{-1}z^k \end{align*} which could be useful when we try to write (1) as telescoping sum.

• We have a representation of reciprocal binomial coefficients via the Beta function:

\begin{align*} \binom{2k}{k}^{-1}&=(2k+1)\int_0^1z^k(1-z)^{k}\,dz\\ &=(2k+1)\int_0^1z^k\sum_{r=0}^k\binom{k}{r}(-z)^r\,dz\\ &=(2k+1)\sum_{r=0}^k\binom{k}{r}(-1)^r\int_0^1z^{k+r}\,dz\\ &=(2k+1)\sum_{r=0}^k\binom{k}{r}\frac{(-1)^r}{k+r+1}\\ &=(2k+1)\sum_{r=0}^k\binom{k}{r}\frac{(-1)^{k-r}}{2k+1-r} \end{align*} which could be useful to transform (1) into a double sum, exchange the sums and try to apply some telescoping.

• We also have the generating function of associated Legendre polynomials (see e.g. Binomial identities by J. Riordan): \begin{align*} \sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n+k}{2k}z^k\right)y^n=\frac{1-y}{1-(2+z)y+y^2} \end{align*}

Regrettably, despite this info I wasn't able to show (1).

Answer 1

We write le LHS term as $$ \eqalign{ & L(n) = \sum\limits_{k = 1}^n {{n \over 4}{{\left( { - 16} \right)^{\,k} } \over {k\left( {n + k} \right)}} \left( \matrix{ n + k \cr 2k \cr} \right)\left( \matrix{ 2k \cr k \cr} \right)^{\, - \,1} } = \cr & = \sum\limits_{1\, \le \,k} {{n \over 4}{{\left( { - 16} \right)^{\,k} } \over {k\left( {n + k} \right)}} \left( \matrix{ n + k \cr n - k \cr} \right)\left( \matrix{ 2k \cr k \cr} \right)^{\, - \,1} } = \cr & = \sum\limits_{0\, \le \,k} { - {{4n\left( { - 16} \right)^{\,k} } \over {\left( {k + 1} \right)\left( {n + k + 1} \right)}} \left( \matrix{ n + k + 1 \cr n - k - 1 \cr} \right)\left( \matrix{ 2k + 2 \cr k + 1 \cr} \right)^{\, - \,1} } = \cr & = \sum\limits_{0\, \le \,k} {T(k,n)} \cr} $$ so to get rid of the upper bound in the sum.

Then we reshape $T(k,n)$ so to render it more manageable by making use of the gamma duplication formula $$ \eqalign{ & T(k,n) = \cr & = {{ - \,4n\left( { - 16} \right)^{\,k} } \over {\left( {k + 1} \right)\left( {n + k + 1} \right)}} \left( \matrix{ n + k + 1 \cr n - k - 1 \cr} \right)\left( \matrix{ 2k + 2 \cr k + 1 \cr} \right)^{\, - \,1} \quad \left| {\,0 \le k \le n - 1} \right.\quad = \cr & = - \,4n{{\Gamma \left( {n + 2 + k} \right)} \over {\left( {n + 1 + k} \right)\Gamma \left( {n - k} \right)\Gamma \left( {2k + 3} \right)}} {{\Gamma \left( {k + 2} \right)^{\,2} } \over {\left( {k + 1} \right)\Gamma \left( {2k + 3} \right)}}\left( { - 16} \right)^{\,k} = \cr & = - \,4n{{\Gamma \left( {n + 1 + k} \right)} \over {\Gamma \left( {n - k} \right)}} {{\Gamma \left( {k + 2} \right)\Gamma \left( {k + 1} \right)} \over {\Gamma \left( {2k + 3} \right)^{\,2} }}\left( { - 16} \right)^{\,k} = \cr & = - \,4n{{\Gamma \left( {n + 1 + k} \right)} \over {\Gamma \left( {n - k} \right)}} {{\Gamma \left( {3/2} \right)^{\,2} \Gamma \left( {k + 2} \right)\Gamma \left( {k + 1} \right)} \over {\Gamma \left( {k + 3/2} \right)^{\,2} \Gamma \left( {k + 2} \right)^{\,2} }}{{\left( { - 16} \right)^{\,k} } \over {4^{\,2\,k + 1} }} = \cr & = - \,n{{\Gamma \left( {n + 1 + k} \right)} \over {\Gamma \left( {n - k} \right)}} {{\Gamma \left( {3/2} \right)^{\,2} \Gamma \left( {k + 1} \right)} \over {\Gamma \left( {k + 3/2} \right)^{\,2} \Gamma \left( {k + 2} \right)}} \left( { - 1} \right)^{\,k} = \cr & = - n^{\,2} {{{{\Gamma \left( {n + 1 + k} \right)} \over {\Gamma \left( {n + 1} \right)}}} \over {{{\Gamma \left( {n - k} \right)} \over {\Gamma \left( n \right)}}}}{1 \over {{{\Gamma \left( {k + 3/2} \right)} \over {\Gamma \left( {3/2} \right)^{\,2} }}^{\,2} }}{{\left( { - 1} \right)^{\,k} } \over {\left( {k + 1} \right)}} = \cr & = - n^{\,2} {{\left( {n + 1} \right)^{\,\overline {\,k\,} } } \over {n^{\,\overline {\, - \,k\,} } }} {1 \over {\left( {3/2} \right)^{\,\overline {\,k\,} } \left( {3/2} \right)^{\,\overline {\,k\,} } }} {{\left( { - 1} \right)^{\,k} } \over {\left( {k + 1} \right)}} = \cr & = - n^{\,2} {{\left( {n - 1} \right)^{\,\underline {\,k\,} } \left( {n + 1} \right)^{\,\overline {\,k\,} } } \over {\left( {3/2} \right)^{\,\overline {\,k\,} } \left( {3/2} \right)^{\,\overline {\,k\,} } }}{{\left( { - 1} \right)^{\,k} } \over {\left( {k + 1} \right)}} = \cr & = - n^{\,2} {{\left( { - n + 1} \right)^{\,\overline {\,k\,} } \left( {n + 1} \right)^{\,\overline {\,k\,} } } \over {\left( {3/2} \right)^{\,\overline {\,k\,} } \left( {3/2} \right)^{\,\overline {\,k\,} } }}{1 \over {\left( {k + 1} \right)}} = \cr & = {{\left( { - n} \right)^{\,\overline {\,k + 1\,} } n^{\,\overline {\,k + 1\,} } } \over {\left( {3/2} \right)^{\,\overline {\,k\,} } \left( {3/2} \right)^{\,\overline {\,k\,} } }}{1 \over {\left( {k + 1} \right)}} \cr} $$

where the single steps should result quite clear.

We do not attempt to go through the Hypergeometric at this point, which looks complicate.
Instead we go and take the Forward Difference in $n$ $$ \Delta _{\,n} L(n) = \sum\limits_{0\, \le \,k} {\Delta _{\,n} T(k,n)} $$ taking advantage of not having the upper bound.

Now $$ \eqalign{ & \Delta _{\,n} \left( {\left( { - n} \right)^{\,\overline {\,k + 1\,} } n^{\,\overline {\,k + 1\,} } } \right) = \cr & = \left( { - n - 1} \right)^{\,\overline {\,k + 1\,} } \left( {n + 1} \right)^{\,\overline {\,k + 1\,} } - \left( { - n} \right)^{\,\overline {\,k + 1\,} } n^{\,\overline {\,k + 1\,} } = \cr & = \left( { - n - 1} \right)\left( { - n} \right)^{\,\overline {\,k\,} } \left( {n + 1} \right)^{\,\overline {\,k\,} } \left( {n + k + 1} \right) - \left( { - n} \right)^{\,\overline {\,k\,} } \left( { - n + k} \right)n\left( {n + 1} \right)^{\,\overline {\,k\,} } = \cr & = - \left( {2n + 1} \right)\left( {k + 1} \right)\left( { - n} \right)^{\,\overline {\,k\,} } \left( {n + 1} \right)^{\,\overline {\,k\,} } \cr} $$ which provides a $(k+1)$ factor, wishfully expected from the Rising Factorials to cancel the disturbing one in the previous derivation

Therefore $$ \eqalign{ & \Delta _{\,n} L(n) = - \left( {2n + 1} \right)\sum\limits_{0\, \le \,k} {{{\left( { - n} \right)^{\,\overline {\,k\,} } \left( {n + 1} \right)^{\,\overline {\,k\,} } 1^{\,\overline {\,k\,} } } \over {\left( {3/2} \right)^{\,\overline {\,k\,} } \left( {3/2} \right)^{\,\overline {\,k\,} } }}{1 \over {k!}}} = \cr & = - \left( {2n + 1} \right){}_3F_{\,2} \left( {\left. {\matrix{ { - n,\;1,\;n + 1} \cr {3/2,3/2} \cr } \;} \right|\;1} \right) \cr} $$ and we are lucky enough that the factors allow to apply the Saalschütz's theorem $$ \eqalign{ & \Delta _{\,n} L(n) = - \left( {2n + 1} \right){{\left( {3/2 - 1} \right)^{\,\overline {\,n\,} } \left( {3/2 - n - 1} \right)^{\,\overline {\,n\,} } } \over {\left( {3/2} \right)^{\,\overline {\,n\,} } \left( {3/2 - 1 - n - 1} \right)^{\,\overline {\,n\,} } }} = \cr & = - \left( {2n + 1} \right){{\left( {1/2} \right)\left( { - 1/2} \right)} \over {\left( {1/2 + n} \right)\left( { - n - 1/2} \right)}} = \cr & = {1 \over {\left( { - 2n - 1} \right)}} \cr} $$

In conclusion, indicating with $R(n)$ the RHS of the identity to demonstrate, we have $$ \left\{ \matrix{ L(0) = R(0) = 0 \hfill \cr \Delta _{\,n} L(n) = \Delta _{\,n} R(n) = - {1 \over {\left( {2n + 1} \right)}} \hfill \cr} \right. $$

and the thesis is proved

Answer 2

Note: There is an interesting integral representation of (1), namely

\begin{align*} \frac{n}{4}\sum_{k=1}^n\frac{(-16)^k}{k(n+k)}\binom{n+k}{2k}\binom{2k}{k}^{-1} \color{blue}{=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{\cos(2nz)-1}{\sin z}\,dz =}\sum_{k=1}^n\frac{1}{1-2k} \end{align*}

The validity of this equality-chain is shown in this MSE post.