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.
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.
The identity under question can be found in the paper S. Boettner, V.H. Moll The integrals in Gradshteyn and Ryzhik. Part 16: Complete elliptic integrals, pages 11-12 https://arxiv.org/abs/1005.2941
The proof given by OP is miraculously similar to the one in this paper, not only in the method used, but also in the choice of words and formatting of equations. By the way, one of the authors V. Moll, is a frequent collaborator of the OP's, and they have written numerous papers that start in the same way The integrals in Gradshteyn and Ryzhik. Part ..., e.g. https://arxiv.org/abs/1004.2440. Coincidences happen, and with some people they happen more often than with others, this is just law of probabilities. No foul play is suspected here. But it is really intriguing to know, how is the identity under question related to digamma function, as OP claims? Unfortunately his proof does not contain any digamma functions. But in the paper the identity naturally came out of some work on the elliptic integral. Please, @T.Amdeberhan, would you share your insight? It is really intriguing to know, how is digamma function related here?
Here is a screenshot of OP's own answer, just in case:
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.
ADDED. The identity under question can be found in the paper S. Boettner, V.H. Moll The integrals in Gradshteyn and Ryzhik. Part 16: Complete elliptic integrals, pages 11-12 https://arxiv.org/abs/1005.2941 Also read on page 14 to find: "Acknowledgments. The authors wish to thank T. Amdeberhan for many comments on a first draft of this paper. In particular, both proofs of Lemma 7.1 are due to him."
