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This came out of some work on the digamma function.

Let $(x)_k=x(x+1)\cdots(x+k-1)$ denote the Pochhammer symbol. Then,

Question. Can you prove/disprove this identity? $$\pmb{\frac{(\frac12)_j^2}{j!^2}}\sum_{i=0}^{j-1}\frac4{2i+1} =\sum_{i=0}^{j-1}\pmb{\frac{(\frac12)_i^2}{i!^2}}\frac1{j-i}.$$

I found this fascinating in view of fact that the factors in bold are able "go in and out" of the sum.

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Here's a sketch of a proof using "creative telescoping."

Let $$T(i,j) = \frac{j!^2}{(\tfrac12)_j^2}\cdot \frac{(\tfrac12)_i^2}{i!^2}\frac1{j-i}.$$ Since the identity holds for $j=1$, it suffices to show that $$\sum_{i=0}^{j} T(i,j+1) -\sum_{i=0}^{j-1}T(i,j)=\frac{4}{2j+1};$$ i.e., that $$T(j,j+1) +\sum_{i=0}^{j-1}\bigl(T(i,j+1) - T(i,j)\bigr) = \frac{4}{2j+1}.\tag{1}$$ But it is easy to verify the indefinite summation $$\sum_{i=0}^{k}\bigl(T(i,j+1) - T(i,j)\bigr) =\frac{j!^2}{(\tfrac32)_j^2}\cdot \frac{(\tfrac32)_k^2}{k!^2}\frac{1}{k-j}$$ from which $(1)$ follows by a straightforward computation.

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  • $\begingroup$ It's not clear how you could "verify the indefinite summation". $\endgroup$ – Lewi_Sol May 31 '17 at 14:37
  • $\begingroup$ To prove that $\sum_{i=0}^k a_i = b_k$ we need only check that $a_0=b_0$ and $b_k - b_{k-1}=a_k$. $\endgroup$ – Ira Gessel May 31 '17 at 16:29
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Let me add an alternative proof I just found.

Using the relation $(-x)_k=(-1)^k(x-k+1)_k$ and $(\frac12)_{n-k}(\frac12-n)_k=(-1)^k(\frac12)_n$, rewrite $$\sum_{i=0}^{j-1}\frac{(\frac12)_i^2}{i!^2}\frac1{j-i} =\sum_{k=0}^{j-1}\frac{(\frac12)_{j-k-1}^2}{(j-k-1)!^2}\frac1{k+1} =\frac{(\frac12)_j^2}{j!^2}\sum_{k=0}^{j-1}\frac{(-j)_{k+1}^2}{(\frac12-j)_{k+1}^2}\frac1{k+1}.$$ The assertion amounts to $\sum_{k=0}^{j-1}\frac{(-j)_{k+1}^2}{(\frac12-j)_{k+1}^2}\frac1{k+1}=\sum_{k=0}^{j-1}\frac4{2k+1}$. Apply $(x)_{k+1}=x(x+1)_k$ so that $$\sum_{k=0}^{j-1}\frac{(-j)_{k+1}^2}{(\frac12-j)_{k+1}^2}\frac1{k+1} =\frac{j^2}{(\frac12-j)^2} \sum_{k=0}^{j-1}\frac{(1-j)_k^2}{(\frac32-j)_k^2}\frac1{k+1} =\frac{j^2}{(\frac12-j)^2} \sum_{k=0}^{j-1}\frac{(1-j)_k^2(1)_k^2}{(\frac32-j)_k^2(2)_kk!}.$$ The RHS is a balanced ${}_4F_3$ series and can be transformed by Bailey tract [Generalized Hypergeometric series, Stechert-Hafner, New York, 1964, p.56] or any classical reference: \begin{align} 4F_3[x,y,z,-m;u,v,w]&=\frac{(v-z)_m(w-z)_m}{(v)_m(w)_m} \times \\ &{}_4F_3[u-x,u-y,z-m;1-v+z-m,1-w+z-m,u]. \end{align} Let $y=z=1, x=1-j, m=j-1, u=v=\frac32-j$ and $w=2$. Then, \begin{align} \sum_{k=0}^{j-1}\frac{(1-j)_k^2(1)_k^2}{(\frac32-j)_k^2(2)_kk!} &={}_4F_3[1,1,1-j,1-j;\frac32-j,\frac32-j,2] \\ &=\frac{(\frac12-j)_{j-1}(1)_{j-1}}{(\frac32-j)_{j-1}(2)_{j-1}} {}_4F_3[1,\frac12,-j+\frac12,1-j;-j+\frac32,\frac32,1-j] \\ &=\frac{2j-1}j\sum_{k=0}^{j-1}\frac{(\frac12)_k(-j+\frac12)_k}{(\frac32)_k(-j+\frac32)_k} \\ &=\frac{(2j-1)^2}j\sum_{k=0}^{j-1}\frac1{(2k+1)(2j-1-2k)} \\ &=\frac{(2j-1)^2}{2j^2}\sum_{k=0}^{j-1}\left[\frac1{2k+1}+\frac1{2j-1-2k} \right] \\ &=\frac{(2j-1)^2}{j^2}\sum_{k=0}^{j-1}\frac1{2k+1}. \end{align} This completes the proof.

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