Solicitud de ayuda para encontrar una expresión de forma cerrada para$\sum_{n=0}^{\infty}\frac{(2n+15)^2}{(2n+11)^3\sqrt{(2n+13)}}$

Question

Por la prueba de comparación, la serie converge. $$\sum_{n=0}^{\infty}\frac{(2n+15)^2}{(2n+11)^3\sqrt{(2n+13)}}=l$$ $$l\simeq0.39483198670640570922670209458869656945532664947383304288146572043505953641$$ (74 dígitos que se muestran)

Me gustaría saber si es posible encontrar una forma cerrada de expresión para $l$.

Editar

He utilizado un PARI/GP función (sumnum) para calcular un valor aproximado de $l$:

 gp > #
    timer = 1 (on)
 gp > \p 74
    realprecision = 77 significant digits (74 digits displayed)
 gp > sumnum(n=0, (2*n+15)^2/(2*n+11)^3/sqrt(2*n+13),sumnuminit([+oo,-1.5]))
 time = 15 ms.
 %1 = 0.39483198670640570922670209458869656945532664947383304288146572043505953641
 gp > sum(k=0,97,binomial(-1/2,k)*(zetahurwitz(k+3/2,11/2)+4*zetahurwitz(k+5/2,11/2)+4*zetahurwitz(k+7/2,11/2)))/2/sqrt(2)
 time = 62 ms.
 %2 = 0.39483198670640570922670209458869656945532664947383304288146572043505953641
 gp > \p 400
    realprecision = 404 significant digits (400 digits displayed)
 gp > sumnum(n=0, (2*n+15)^2/(2*n+11)^3/sqrt(2*n+13),sumnuminit([+oo,-1.5]))
 time = 1,078 ms.
 %3 = 0.3948319867064057092267020945886965694553266494738330428814657204350595364122729251535099721677148100865638904442238183504239633022203492259575689525638233484603909542517059855919506733374353926795281971622814159109637670427399334579124116308394274900167149611375839262362929064148759669151317574581119034347857659144521214162698550204609725028401793070583608454779328237780415473232520384318559950198
 gp > sum(k=0,537,binomial(-1/2,k)*(zetahurwitz(k+3/2,11/2)+4*zetahurwitz(k+5/2,11/2)+4*zetahurwitz(k+7/2,11/2)))/2/sqrt(2)
 time = 3,532 ms.
 %4 = 0.3948319867064057092267020945886965694553266494738330428814657204350595364122729251535099721677148100865638904442238183504239633022203492259575689525638233484603909542517059855919506733374353926795281971622814159109637670427399334579124116308394274900167149611375839262362929064148759669151317574581119034347857659144521214162698550204609725028401793070583608454779328237780415473232520384318559950198

Answer 1

No de una forma cerrada, pero una muy buena aproximación de la misma.

La reescritura como $$a_n=\frac{(2n+15)^2}{(2n+11)^3\sqrt{2n+13}}=\frac{(2n+15)^2}{(2n+11)^3\sqrt{2n+11+2}}=\frac{(2n+15)^2}{(2n+11)^{\frac 72}\sqrt{1+\frac 2{2n+11}}}$$ , podemos escribir $$\frac 1{\sqrt{1+\frac 2{2n+11}}}=\sum_{k=0}^\infty \binom{-\frac{1}{2}}{k}\frac{2^k}{(2n+11)^k}$$ y, a continuación, la cara sumatorias de los términos $$S_k=\sum_{n=0}^\infty\frac{(2n+15)^2}{(2n+11)^{k+\frac 72}}=2^{-k-\frac{3}{2}} \left(\zeta \left(k+\frac{3}{2},\frac{11}{2}\right)+4 \zeta \left(k+\frac{5}{2},\frac{11}{2}\right)+4 \zeta \left(k+\frac{7}{2},\frac{11}{2}\right)\right)$$ y, a continuación, por la que se considera de suma $$\Sigma=\sum_{n=0}^\infty \frac{(2n+15)^2}{(2n+11)^3\sqrt{2n+13}}$$ Así, las sumas parciales $$\Sigma_p=\frac 1{2\sqrt 2}\sum_{k=0}^p \binom{-\frac{1}{2}}{k} \left(\zeta \left(k+\frac{3}{2},\frac{11}{2}\right)+4 \zeta \left(k+\frac{5}{2},\frac{11}{2}\right)+4 \zeta \left(k+\frac{7}{2},\frac{11}{2}\right)\right)$$ Computación $$\left( \begin{array}{cc} p & \Sigma_p \\ 5 & 0.39483148307398830445048206504782536496698034745960346801382376024 \\ 10 & 0.39483198676616324735777911451436673248113391244073226133701558847 \\ 15 & 0.39483198670639658527502248284985700122310258168712958250026452734 \\ 20 & 0.39483198670640571076206255275333399438974388443990146144695977886 \\ 25 & 0.39483198670640570922643131344152894240169346679460070946980052257 \\ 30 & 0.39483198670640570922670214360763624644791582367286943018650562365 \\ 35 & 0.39483198670640570922670209457967764179263574295643280410747243813 \\ 40 & 0.39483198670640570922670209458869824722904384200792304484978282169 \\ 45 & 0.39483198670640570922670209458869656914069678422236918330533995116 \\ 50 & 0.39483198670640570922670209458869656945538601634396101492730552390 \\ 55 & 0.39483198670640570922670209458869656945532663821679612114844784572 \\ 60 & 0.39483198670640570922670209458869656945532664947597620995355367822 \\ 65 & 0.39483198670640570922670209458869656945532664947383263347402134364 \\ 70 & 0.39483198670640570922670209458869656945532664947383304295989988132 \\ 75 & 0.39483198670640570922670209458869656945532664947383304288145065669 \\ 80 & 0.39483198670640570922670209458869656945532664947383304288146572333 \\ 85 & 0.39483198670640570922670209458869656945532664947383304288146572043 \\ 90 & 0.39483198670640570922670209458869656945532664947383304288146572044 \\ 95 & 0.39483198670640570922670209458869656945532664947383304288146572044 \\ 100 & 0.39483198670640570922670209458869656945532664947383304288146572044 \end{array} \right)$$

Editar

Con el fin de saber cuántos términos tiene que ser añadido para $p$ dígitos significativos, ya que nos enfrentamos a un alternatin de la serie, considere la posibilidad de $$a_k=\frac 1{2\sqrt 2}\binom{-\frac{1}{2}}{k} \left(\zeta \left(k+\frac{3}{2},\frac{11}{2}\right)+4 \zeta \left(k+\frac{5}{2},\frac{11}{2}\right)+4 \zeta \left(k+\frac{7}{2},\frac{11}{2}\right)\right)$$ A quick and dirty linear regression (built for $10 \leq k \leq 1000$ by steps of $10$) muestra que (y esto es casi un ajuste perfecto) $$\log_{10}(|a_k|)=-0.740989\, k-2.51445$$ So, for $p$ exact digits, we need to sum up $\lceil 1.35\, p -3.40 \rceil$ terms (just as reflected by the values in the above table). For $74$ digits as given in the post, then $k=97$ which has been verified. Using the original summation, I suppose that billions of terms have been added (summing from $n=0$ to $n=10^9$ leading to $0.39481$). This look normal since, for large values of $n$, $\frac{a_{n+1}}{a_n}=1-\frac{3}{2 n}+O\left(\frac{1}{n^2}\right)$ que es extremadamente lento.