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.



Identity with Pochhammer and harmonic numbers
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.


