Proving that $ g_N(\frac{x}{2}) + g_N(\frac{x+1}{2}) = 2g_{2N}(x) + \frac{2}{x+2N+1}$

Question

I was asked to prove that

$$ g_N(\frac{x}{2}) + g_N(\frac{x+1}{2}) = 2g_{2N}(x) + \frac{2}{x+2N+1}$$

where

$$g_N(x) = \sum_{n=-N}^{N}\frac{1}{x+n}$$

My teacher swears it is indeed true, but I feel like it can't be, here is my calculation so far:

$$ g_N(\frac{x}{2}) + g_N(\frac{x+1}{2}) = \sum_{n=-N}^{N}\frac{1}{\frac{x}{2}+n} + \sum_{n=-N}^{N}\frac{1}{\frac{x+1}{2}+n} = \sum_{n=-N}^{N}\frac{1}{\frac{x}{2}+n} + \sum_{n=-N}^{N}\frac{1}{\frac{x}{2}+\frac{1}{2}+n} = 2\sum_{n=-N}^{N}\frac{1}{x+2n} + 2\sum_{n=-N}^{N}\frac{1}{x+2n+1} = 2\sum_{n=-2N}^{2N}\frac{1}{x+n} + 2\sum_{n=-2N}^{2N}\frac{1}{x+n+1} = 2\sum_{n=-2N}^{2N}\frac{1}{x+n} + 2\sum_{n=-2N+1}^{2N+1}\frac{1}{x+n} = 2\sum_{n=-2N}^{2N}\frac{1}{x+n} + 2\sum_{n=-2N}^{2N}\frac{1}{x+n} + \frac{2}{x+2N+1} - \frac{2}{x-2N} = 4\sum_{n=-2N}^{2N}\frac{1}{x+n} + \frac{2}{x+2N+1} - \frac{2}{x-2N}$$

I have no idea how to achieve the result. Reversing the calculation (getting the left side of the equation from the right side didn't work either for me). Do you have an idea how to prove that?

Answer 1

Observe that $$\sum_{n=-N}^N f(2n)+ \sum_{n=-N}^N f(2n+1)=\sum_{n=-2N}^{2N+1}f(n)=\sum_{n=-2N}^{2N}f(n)\;+f(2N+1).$$ Hence you should come up with $$ 2\sum_{n=N}^N\frac1{x+2n}+2\sum_{n=N}^N\frac1{x+2n+1}=2\sum_{n=-2N}^{2N}\frac1{x+n}\;+\frac2{x+2N+1}.$$

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If transformations and index manipulations as above are too confusing, you can always try good old induction (it is worth a try because the sum is essentially defined by a recursion):

We have $$g_0(x)=\frac1{x} $$ and for $n\ge1$, $$g_{n}(x)=g_{n-1}(x)+\frac1{x+n}+\frac1{x-n}.$$ Hence for $N=0$, we have $$ g_0(\tfrac x2)+g_0(\tfrac{x+1}2)=\frac2{x}+\frac2{x+1}=2g_0(x)+\frac2{x+2\cdot 0+1}$$ as desired. See if you can complete the cmputation for $N>0$ (using the induction hypothesis), $$\begin{align}g_N(\tfrac x2)+g_N(\tfrac{x+1}2) &=g_{N-1}(\tfrac x2)+\frac1{\frac x2+N}+\frac1{\frac x2-N}+g_{N-1}(\tfrac{x+1}{2})+\frac1{\frac {x+1}2+N}+\frac1{\frac {x+1}2-N}\\ &=2g_{2N-2}(x)+\frac2{x+2N-1}+\frac1{\frac x2+N}+\frac1{\frac x2-N}+\frac1{\frac {x+1}2+N}+\frac1{\frac {x+1}2-N}\\ &=\qquad\vdots\quad?\\&=2g_{2N}(x)+\frac2{x+2N+1}\end{align}$$

Answer 2

This exercise can be used to prove the duplication formula for the $\Gamma$ function by proving the duplication formula for the $\psi$ function for first. $g_N(x)$ is a meromorphic function with a simple pole with unit residue at every $x\in[-N,N]$. It follows that $g_N(x/2)$ is a meromorphic function with a simple pole with residue $2$ at every even $x\in[-2N,2N]$ and $g_N((1+x)/2)$ is a meromorphic function with a simple pole with residue $2$ at every odd $x\in[-2N-1,2N-1]$.
As a consequence, both $$ g_N\left(\frac{x}{2}\right)+g_N\left(\frac{1+x}{2}\right),\qquad 2\,g_{2N}(x)+\frac{2}{x+2N+1} $$ are meromorphic functions with a pole with residue $2$ at every $x\in[-2N-1,2N]$.
In particular, they are the same function. Have a look at Herglotz trick.