Simplify $\sum_{l=0}^\infty \sum_{r=0}^\infty\frac{\Gamma(L+r-2q)}{\Gamma(L+r-1+2q)} \frac{\Gamma(L+r+l-1+2q)}{\Gamma(L+r+l+2)}\frac{r+1}{r+l+2}$

Question

Simplify $\sum_{l=0}^\infty \sum_{r=0}^\infty\frac{\Gamma(L+r-2q)}{\Gamma(L+r-1+2q)} \frac{\Gamma(L+r+l-1+2q)}{\Gamma(L+r+l+2)}\frac{r+1}{r+l+2}$

863 Views Asked by Bumbble Comm At 27 Mar 2026 - 1:00

This question is a continuation of this post.

Let $r,l,L\geq 1$ be integers. Assume that $q\in [0,1]$ is a real number.

The authors obtained the following equation $36$ in their paper (I express in summation forms).

$$\sum_{r=1}^\infty \sum_{l=1}^\infty \frac{r}{r+l} 4(1-q)q \frac{Γ(L)}{Γ(L − 2p)} \frac{Γ(L + l − 1 − 2p)}{Γ(L + l − 2q)} \frac{Γ(L + l + r − 1 − 2q)}{ Γ(L + l + r) } = $$ $$ 1 - \frac{2q^2}{(1-2q)^2} - \frac{2\pi q(1-q)}{(1-2q)(3-4q)} \cot(2\pi q).$$ Note that the above formula are equal in the sense of asymptotically.

In the midst of obtaining the above formula, I stuck at the following.

Question: Is it possible to express $$\sum_{l=0}^\infty \sum_{r=0}^\infty\frac{\Gamma(L+r-2q)}{\Gamma(L+r-1+2q)} \frac{\Gamma(L+r+l-1+2q)}{\Gamma(L+r+l+2)}\frac{r+1}{r+l+2}$$ in closed form independent of summations (possibly in terms of Gamma function)?

I tried to express above double summations using Appell's series. However, I stuck at $$\frac{\Gamma(L-2q)}{\Gamma(L+2)} \sum_{l=0}^\infty \sum_{r=0}^\infty \frac{(L-1+2q)_{l+r}}{(L+2)_{l+r}} \frac{(L-2q)_r}{(L-1+2q)_r} \frac{(1)_l}{l!r!} \frac{(2)_r}{(r+l+2)}.$$

In particular, can we express $$\frac{(L-2q)_r}{(L-1+2q)_r (r+l+2)}$$ to be something which allows us to use Appell series?

I also tried to evaluate the summation over $l$, that is, $$\sum_{l=0}^\infty \frac{\Gamma(L+r-1+2q+l)}{\Gamma(L+r+2+l)(r+l+2)},$$ and I arrive at $$_3F_2(L+r-1+2q,1,r+2; L+r+2,r+3;1)\frac{\Gamma(L+r-1+2q)}{\Gamma(L+r+2)(r+2)}.$$ However, I could not find any formula to evaluate $_3F_2(L+r-1+2q,1,r+2; L+r+2,r+3;1)$ in Wolfram Alpha.

Updated (2 Nov 2018): @Nikos Bagis uses the Mathematica 10 to obtain a closed form for my question. However, I would like to have a detailed calculations on how to obtain the answer.

Original Q&A

There are 2 best solutions below

Bumbble Comm On 02 Nov 2018 - 11:57

The sum to find can be written in the two equivalent forms, as demonstrated in the precedent post. $$ \eqalign{ & S(L,q) = \sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {{{\Gamma (L + r - 2q)} \over {\Gamma (L + r - 1 + 2q)}}{{\Gamma (L + r + l - 1 + 2q)} \over {\Gamma (L + r + l + 2)}}{{r + 1} \over {r + l + 2}}} } = \cr & = \sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {\left( {r + 1} \right)\left( {L + r - 1 + 2q} \right)^{\,\overline {\, - \left( { 4q-1} \right)\,} } \left( {L + r + l + 2} \right)^{\,\overline {\, - 3 + 2q\,} } {1 \over {r + l + 2}}} } \cr & = \sum\limits_{s = 0}^\infty {{1 \over {s + 2}}\left( {L + s + 2} \right)^{\,\overline {\, - 3 + 2q\,} } \sum\limits_{r = 0}^s {\left( {r + 1} \right)\left( {L + r - 1 + 2q} \right)^{\,\overline {\, - \left( { 4q-1} \right)\,} } } } \cr} $$

Now, using the composition formula for the Rising Factorial $$ x^{\,\overline {\,a + b\,} } = x^{\,\overline {\,a\,} } \left( {x + a} \right)^{\,\overline {\,b\,} } $$ we obtain two fundamental blocks that we need for proceeding $$ \eqalign{ & \left( {r + 1} \right)\left( {r + 1 + a} \right)^{\,\overline {\,1 - b\,} } = {{\left( {r + 1} \right)^{\,\overline {\,1\,} } \left( {r + 2} \right)^{\,\overline {\,a - 1\,} } \left( {r + 1 + a} \right)^{\,\overline {\,1 - b\,} } } \over {\left( {r + 2} \right)^{\,\overline {\,a - 1\,} } }} = \cr & = {{\left( {r + 1} \right)^{\,\overline {\,a + 1 - b\,} } } \over {\left( {r + 2} \right)^{\,\overline {\,a - 1\,} } }} = {{1^{\,\overline {\,r + 1\,} } } \over {1^{\,\overline {\,r\,} } }}{{1^{\,\overline {\,r\,} } \left( {r + 1} \right)^{\,\overline {\,a + 1 - b\,} } } \over {1^{\,\overline {\,r + 1\,} } \left( {r + 2} \right)^{\,\overline {\,a - 1\,} } }} = {{2^{\,\overline {\,r\,} } } \over {1^{\,\overline {\,r\,} } }}{{1^{\,\overline {\,r + a + 1 - b\,} } } \over {1^{\,\overline {\,r + a\,} } }} = \cr & = {{2^{\,\overline {\,r\,} } } \over {1^{\,\overline {\,r\,} } }}{{1^{\,\overline {\,a + 1 - b\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } } \over {1^{\,\overline {\,a\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }} = \left( {a + 1} \right)^{\,\overline {\,1 - b\,} } {{2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } } \over {1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }} \cr} $$ and $$ \eqalign{ & {1 \over {y + 2}}\left( {y + 2 + c} \right)^{\,\overline {\,d\,} } = {{\left( {y + 3} \right)^{\,\overline {\,c - 1\,} } \left( {y + 2 + c} \right)^{\,\overline {\, - d\,} } } \over {\left( {y + 2} \right)^{\,\overline {\,1\,} } \left( {y + 3} \right)^{\,\overline {\,c - 1\,} } }} = \cr & = {{\left( {y + 3} \right)^{\,\overline {\,c - d - 1\,} } } \over {\left( {y + 2} \right)^{\,\overline {\,c\,} } }} = {{1^{\,\overline {\,y + 1\,} } } \over {1^{\,\overline {\,y + 2\,} } }}{{1^{\,\overline {\,y + 2\,} } \left( {y + 3} \right)^{\,\overline {\,c - d - 1\,} } } \over {1^{\,\overline {\,y + 1\,} } \left( {y + 2} \right)^{\,\overline {\,c\,} } }} = \cr & = {{1^{\,\overline {\,y + 1\,} } } \over {1^{\,\overline {\,y + 2\,} } }}{{1^{\,\overline {\,y + 1 + c - d\,} } } \over {1^{\,\overline {\,y + 1 + c\,} } }} = {{1^{\,\overline {\,1\,} } 2^{\,\overline {\,y\,} } } \over {1^{\,\overline {\,2\,} } 3^{\,\overline {\,y\,} } }} {{1^{\,\overline {\,1 + c - d\,} } \left( {2 + c - d} \right)^{\,\overline {\,y\,} } } \over {1^{\,\overline {\,1 + c\,} } \left( {2 + c} \right)^{\,\overline {\,y\,} } }} = \cr & = {{1^{\,\overline {\,1\,} } 1^{\,\overline {\,1 + c - d\,} } } \over {1^{\,\overline {\,2\,} } 1^{\,\overline {\,1 + c\,} } }} {{2^{\,\overline {\,y\,} } \left( {2 + c - d} \right)^{\,\overline {\,y\,} } } \over {3^{\,\overline {\,y\,} } \left( {2 + c} \right)^{\,\overline {\,y\,} } }} = \cr & = {{\left( {c + 2} \right)^{\,\overline {\, - d\,} } } \over 2}{{2^{\,\overline {\,y\,} } \left( {2 + c - d} \right)^{\,\overline {\,y\,} } } \over {3^{\,\overline {\,y\,} } \left( {2 + c} \right)^{\,\overline {\,y\,} } }} \cr} $$

1) the $(r,l)$ version

Applying the above to the sum in $(r,l)$ $$ \bbox[lightyellow] { \eqalign{ & \left\{ \matrix{ \matrix{ {a = L - 2 + 2q} \hfill & {b = \,4q} \hfill \cr {c = L} \hfill & {d = 3 - 2q\,} \hfill \cr } \hfill \cr A = {1 \over 2}\left( {c + 2} \right)^{\,\overline {\, - d\,} } \left( {a + 1} \right)^{\,\overline {\,1 - b\,} } = {1 \over 2}\left( {L + 2} \right)^{\,\overline {\, - 2 - 2q\,} } \hfill \cr} \right. \cr & S(L,q) = \sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {\left( {r + 1} \right)\left( {r + 1 + L - 2 + 2q} \right)^{\,\overline {\,1 - 4q\,} } \left( {L + r + l + 2} \right)^{\,\overline {\, - \left( {3 - 2q} \right)\,} } {1 \over {r + l + 2}}} } = \cr & = A\;\sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {{{2^{\,\overline {\,r + l\,} } \left( {2 + c - d} \right)^{\,\overline {\,r + l\,} } } \over {3^{\,\overline {\,r + l\,} } \left( {2 + c} \right)^{\,\overline {\,r + l\,} } }}{{2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } } \over {1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} } \cr} } \tag{1.1}$$

Now, the above, once rewritten as $$ \bbox[lightyellow] { \eqalign{ & S(L,q) = A\;\sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {{{2^{\,\overline {\,r + l\,} } \left( {2 + c - d} \right)^{\,\overline {\,r + l\,} } } \over {3^{\,\overline {\,r + l\,} } \left( {2 + c} \right)^{\,\overline {\,r + l\,} } }} {{2^{\,\overline {\,r\,} } 1^{\,\overline {\,l\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } 1^{\,\overline {\,l\,} } } \over {\left( {a + 1} \right)^{\,\overline {\,r\,} } 1^{\,\overline {\,l\,} } }}} } {{x^{\,r} } \over {r!}}{{y^{\,l} } \over {l!}} = \cr & = A\,F\left( {\matrix{ 2 \cr 3 \cr 2 \cr 2 \cr } \,\left| {\,\matrix{ {2;\left( {2 + c - d} \right)} \cr {2,\,1\;;\;\left( {a + 2 - b} \right),1\;} \cr {3;\left( {2 + c} \right)} \cr {\;\left( {a + 1} \right),1} \cr } \,} \right|x,y} \right)\quad \quad \left| {\;x = y = 1} \right. \cr} } \tag{1.2}$$

tells us that the sum is a

Kampé de Fériet function computed at $x=y=1$.

Looking on the web one can find various specialized papers dealing with the expansion of Kampé de Fériet function function in terms of Hypergeometric functions. So the expansion found by @Nikos Bagis well enters into the scene. (re. to e.g. this paper and to this and this).

Also interesting is the connection with the "Extended Beta" integral representation (re. for instance to this and this paper, and specially this), which relates with the integral representation hinted in my answer to the previous post

2) the $(r,s)$ truncated sum version

Restarting from (1.1) above, and writing $S(L,q)$ in the truncated way we get $$ \bbox[lightyellow] { \eqalign{ & S(L,q) = \sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {\left( {r + 1} \right)\left( {r + 1 + L - 2 + 2q} \right)^{\,\overline {\,1 - 4q\,} } \left( {L + r + l + 2} \right)^{\,\overline {\, - \left( {3 - 2q} \right)\,} } {1 \over {r + l + 2}}} } = \cr & = A\;\sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {{{2^{\,\overline {\,r + l\,} } \left( {2 + c - d} \right)^{\,\overline {\,r + l\,} } } \over {3^{\,\overline {\,r + l\,} } \left( {2 + c} \right)^{\,\overline {\,r + l\,} } }}{{2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } } \over {1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} } = \cr & = A\;\sum\limits_{s = r + l = 0}^\infty {{{2^{\,\overline {\,s\,} } \left( {2 + c - d} \right)^{\,\overline {\,s\,} } } \over {3^{\,\overline {\,s\,} } \left( {2 + c} \right)^{\,\overline {\,s\,} } }}\sum\limits_{r = 0}^s {{{2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } } \over {1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} } \cr} } \tag{2.1}$$

In order to reduce it, exactly or asymptotically, into more manageable terms, I am currently exploring the possibilities offered by some possible approaches that I could individuate up to now.

2.a) Product of two Hypergeometric A truncated Hypergeometric series can be transformed into the limit of a standard one, by adding two additional parameters $$ \eqalign{ & \sum\limits_{r = 0}^s {{{2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } } \over {1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \sum\limits_{r = 0}^\infty {{{\left( { - s} \right)^{\,\overline {\,r\,} } 2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } } \over {\left( { - s + \varepsilon } \right)^{\,\overline {\,r\,} } 1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} = \cr & = \mathop {\lim }\limits_{\varepsilon \, \to \,0} \sum\limits_{r = 0}^\infty {{{\left( { - s} \right)^{\,\overline {\,\varepsilon \,} } 2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } } \over {\left( { - s + r} \right)^{\,\overline {\,\varepsilon \,} } 1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} \cr} $$ (re. to the renowned "Concrete Mathematics" - Ch. 5 - ex. 34)
which takes advantage of that $\left( { - s} \right)^{\,\overline {\,r\,} }$ becomes null for non negative integers $s < r$.
This gives us a summand product of two hypergeometric terms, but with one fraction containing both parameters.
Since $$ {{\left( { - s} \right)^{\,\overline {\,\varepsilon \,} } } \over {\left( { - s + r} \right)^{\,\overline {\,\varepsilon \,} } }} = {{\Gamma \left( { - s + \varepsilon } \right)\Gamma \left( { - s + r} \right)} \over {\Gamma \left( { - s} \right)\Gamma \left( { - s + r + \varepsilon } \right)}} = {{{\rm B}\left( { - s + r,\varepsilon } \right)} \over {{\rm B}\left( { - s,\varepsilon } \right)}} $$ there might be a possibility to expand it in some interesting way.

2.b) series expression of the truncated Hypergeometric

Trying to apply the formulas provided in this work

**Bumbble Comm** · Accepted Answer

... $$ \sum^{\infty}_{l=0}\sum^{\infty}_{r=0}\frac{\Gamma(L+r-2q)\Gamma(L+r+l-1+2q)}{\Gamma(L+r-1+2q)\Gamma(L+r+l+2)}\frac{r+1}{r+l+2}\tag 0 $$

I have an answer but I use Mathematica 10 for it. Set $$ A_0(L,q,x):=\sum^{\infty}_{l=0}\sum^{\infty}_{r=0}\frac{\Gamma(L+r-2q)\Gamma(L+r+l-1+2q)}{\Gamma(L+r-1+2q)\Gamma(L+r+l+2)}(r+1)x^{l+r+1}. $$ Then using Mathematica evaluate $A_0(L,q,x)$ and FullSimplify it. You will get result. Then integrate the result $$ \int^{1}_{0}A_0(L,q,t)dt $$ and FullSimplify it. You get a function of $\cot$, $\Gamma$, ${}_2F_1$ and ${}_3F_2$ functions. But you have to wait a while.

The function ${}_2F_1$ is computable with elementary functions. It is actualy $$ {}_2F_1(1,-1+L-2q;L;1)=\frac{L-1}{2q} $$ For the ${}_3F_2$, you have to evaluate $$ {}_3F_2(1,1+L-2q,-1+L+2q;2+L,L+2q;1),\tag 1 $$ $$ {}_3F_2(2,2+L-2q,L+2q;3+L,1+L+2q;1)\tag 2 $$ and $$ {}_3F_2(3,3+L-2q,1+L+2q;4+L,2+L+2q;1).\tag 3 $$ Note that these are the zero, first and second derivatives of $$ {}_3F_2(1,1+L-2q,-1+L+2q;2+L,L+2q;x)\tag 4 $$ at $x=1$. So actualy, you have to compute $(4)$, or if you prefere the values $(1),(2),(3)$. The function $(4)$ also referes to the problem of finding $$ \sum^{\infty}_{n=0}\frac{\Gamma(n+a)\Gamma(n+b)}{\Gamma(n+c)\Gamma(n+d)}x^n $$ So the final answer is

$\ldots$

CONTINUED

Assume that $$ F_1(L,q,x):=\sum^{\infty}_{l=0}\sum^{\infty}_{r=0}\frac{\Gamma(L+r-2q)\Gamma(L+r+l-1+2q)}{\Gamma(L+r-1+2q)\Gamma(L+r+l+2)}x^lx^r. $$ Then $$ F_1(L,q,x)=-\frac{\Gamma(L-2q)(L-2q-1)}{4q-2}{}_2\widetilde{F}_1\left(1,L-2q;L+2;x\right)+ $$ $$ \frac{(L+2q-2)\Gamma(L-2q)}{4q-2}{}_2\widetilde{F}_1\left(1,L+2q-1;L+2;x\right)- $$ $$ -\frac{\Gamma(L-2q)}{4q-2}{}_2\widetilde{F}_1\left(2,L-2q;L+2;x\right), $$ where $$ {}_p\widetilde{F}_q\left(a_1,a_2,\ldots,a_p;b_1,b_2,\ldots,b_q;z\right):= $$ $$ {}_pF_q\left(a_1,a_2,\ldots,a_p;b_1,b_2,\ldots,b_q;z\right)/(\Gamma(b_1)\Gamma(b_2)\ldots\Gamma(b_q)) $$ is the regularized hypergeometric function.

Set now $$ F_2(L,q,x):=\sum^{\infty}_{l=0}\sum^{\infty}_{r=0}\frac{\Gamma(L+r-2q) \Gamma(L+r+l-1+2q)}{\Gamma(L+r-1+2q) \Gamma(L+r+l+2)}rx^lx^r. $$ Then $$ F_2(L,q,x)= $$ $$ =\frac{x\Gamma(L-2q+1)(2+L-6q-4Lq+8q^2)}{2(2q-1)(4q-3)\Gamma(L+3)}{}_2F_1\left(2,L-2q+1;L+3;x\right)- $$ $$ -\frac{\Gamma(L-2q+1)(L+2q-2)}{2(2q-1)(4q-3)\Gamma(L+2)}{}_2F_1(1,L-2q;L+2;x)+ $$ $$ +\frac{\Gamma(L-2q+1)(L+2q-2)}{2(2q-1)(4q-3)\Gamma(L+2)}{}_2F_1(1,L+2q-1;L+2;x)- $$ $$ -x\frac{\Gamma(L-2q+1)}{(4q-3)\Gamma(L+3)} {}_3F_2\left(2,2,L-2q+1;1,L+3;x\right). $$ But $$ F_3(L,q,x) =\sum^{\infty}_{l=0}\sum^{\infty}_{r=0}\frac{\Gamma(L+r-2q) \Gamma(L+r+l-1+2q)}{\Gamma(L+r-1+2q) \Gamma(L+r+l+2)}(r+1)x^lx^r= $$ $$ =F_1(L,q,x)+F_2(L,q,x) $$ and $$ S=S(L,q)=\int^{1}_{0}x\cdot F_3(L,q,x)dx $$ But also $$ k_{11}(b,c):=\int^{1}_{0}x\cdot { }_2F_1(1,b;c;x)dx= $$ $$ =\frac{(c-1)}{(b-1)(b-2)}\left[2-b+(c-2)\psi(c-2)-(c-2)\psi(c-b)\right], $$ where $Re(b)<Re(c)$. $$ k_{12}(b,c):=\int^{1}_{0}x\cdot {}_2F_1(2,b;c;x)dx= $$ $$ =\frac{(c-1)(c-2)}{(b-1)(b-2)}\left[-1+{}_2F_1(1,b-2;c-2;1)-\psi(c-2)+\psi(c-b)\right] $$ $$ k_{21}(b,c):=\int^{1}_{0}x^2\cdot { }_2F_1(1,b;c;x)dx =\frac{(c-1)}{2(b-1)(b-2)(b-3)}\times $$ $$ \times\left[-(b-3)(2c+b-6)+2(c-3)(c-2)\psi(c-3)-2(c-3)(c-2)\psi(c-b)\right], $$ $$ k_{22}(b,c):=\int^{1}_{0}x^2\cdot { }_2F_1(2,b;c;x)dx =\frac{(c-1)(c-2)}{(b-1)(b-2)(b-3)}\times $$ $$ \times\left[b-c+(c-3){}_2F_1(1,b-3;c-3;1)-2(c-3)\psi(c-3)+2(c-3)\psi(c-b)\right], $$ and $$ M(b,c):=\int^{1}_{0}x^2\cdot { }_3F_2(2,2,b;1,c;x)dx= $$ $$ =-\frac{18}{(b-3)(b-2)(b-1)}-\frac{2b}{(b-3)(b-2)(b-1)}+ $$ $$ \frac{(c-4)(c-3)(c-2)(c-1)}{(b-1)(b-2)(b-c+1)(b-c+2)} +\frac{35c}{(b-1)(b-2)(b-3)}+ $$ $$ +\frac{3bc}{(b-1)(b-2)(b-3)}-\frac{21c^2}{(b-1)(b-2)(b-3)} -\frac{bc^2}{(b-1)(b-2)(b-3)}+ $$ $$ +\frac{4c^3}{(b-1)(b-2)(b-3)}-\frac{4(c-1)(c-2)(c-3)(c-4)}{(b-1)(b-2)(b-3)(c-b-1)}+ $$ $$ +\frac{4(-6+11c-6c^2+c^3)\psi(c-3)}{(b-1)(b-2)(b-3)}-\frac{4(-6+11c-6c^2+c^3)\psi(b-c)}{(b-1)(b-2)(b-3)}, $$ where $2+Re(b)<Re(c)$.

Hence the value of the double sum (tag $(0)$) is $$ S=S(L,q)=-\frac{\Gamma(L-2q)(L-2q-1)}{(4q-2)\Gamma(L+2)} k_{11}(L-2q,L+2)+ $$ $$ +\frac{(L+2q-2)\Gamma(L-2q)}{(4q-2)\Gamma(L+2)}k_{11}(L+2q-1,L+2)- $$ $$ -\frac{\Gamma(L-2q)}{(4q-2)\Gamma(L+2)}k_{12}(L-2q,L+2)+ $$ $$ +\frac{\Gamma(L-2q+1)(2+L-6q-4Lq+8q^2)}{2(2q-1)(4q-3)\Gamma(L+3)}k_{22}(L-2q+1,L+3)- $$ $$ -\frac{\Gamma(L-2q+1)(L+2q-2)}{2(2q-1)(4q-3)\Gamma(L+2)}k_{11}(L-2q,L+2)+ $$ $$ +\frac{\Gamma(L-2q+1)(L+2q-2)}{2(2q-1)(4q-3)\Gamma(L+2)}k_{11}(L+2q-1,L+2)- $$ $$ -\frac{\Gamma(L-2q+1)}{(4q-3)\Gamma(L+3)}M(L-2q+1,L+3). $$ Where $$ \psi(x):=\frac{\Gamma'(x)}{\Gamma(x)}, $$ where $\Gamma(x)$ is the classical Euler's Gamma function.

After simplification we get $$ S=S(L,q)=\frac{(-2 q (q (4 q-11)+8)-2 \pi (q-1) q (2 q-1) \cot (2 \pi q)+3) \Gamma (L-2 q)}{4 (1-2 q)^2 (q-1) q (4 q-3) \Gamma (L)},\tag 5 $$ with the condition $L\in \textbf{R}$ and $0<q<\frac{3}{2}$.

When $q=3/2-1/t$, $t>>1$ and

i) $L=t^a$, $a=1/3$, then $\lim_{t\rightarrow\infty}S(L,q)=1/24$.

ii) $L=t^a$, $a>1/3$, $\lim_{t\rightarrow\infty}S(L,q)=0$

iii) $L=t^a$, $a<1/3$, $\lim_{t\rightarrow\infty}S(L,q)=\infty$

Also for $q\rightarrow 0+$, then if $L\in\{0,-1,-2,\ldots\}$, we get $\lim_{t\rightarrow\infty}S(L,q)=0$.

If $q\rightarrow 0+$ and $L>0$, then $\lim_{q\rightarrow 0+}S(L,q)=\infty$.

Simplify $\sum_{l=0}^\infty \sum_{r=0}^\infty\frac{\Gamma(L+r-2q)}{\Gamma(L+r-1+2q)} \frac{\Gamma(L+r+l-1+2q)}{\Gamma(L+r+l+2)}\frac{r+1}{r+l+2}$

There are 2 best solutions below

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