Una serie infinita en polygamma función

Question

Estoy interesado en $2$ cosas: $1)$ si estás acostumbrado a esas series, y cuando se reunió antes de dicha serie y $2)$ las herramientas que le gustaría emplear $3)$ no me piden una solución.

Calcular en forma cerrada $$\sum _{n=1}^{\infty } (24 \psi ^{(-3)}(n)-24 \psi ^{(-3)}(n+1)+24 \psi ^{(-2)}(n+1)-12 \text{log$\Gamma $}(n+1)+4 \psi ^{(0)}(n+1)-\psi ^{(1)}(n+1))$$

EDIT: me conjetura de la hermosa forma cerrada $$36 \log (A)+3 \gamma+\zeta (2) -6 \log (2 \pi )-\frac{3 \zeta (3)}{2 \zeta (2)}$$

Aquí hay otro ejemplo con conjuctured forma cerrada

$$\sum _{n=1}^{\infty } (90 \text{log$\Gamma $}(n+1)-3628800 \psi ^{(-9)}(n)+3628800 \psi ^{(-9)}(n+1)-3628800 \psi ^{(-8)}(n+1)+1814400 \psi ^{(-7)}(n+1)-604800 \psi ^{(-6)}(n+1)+151200 \psi ^{(-5)}(n+1)-30240 \psi ^{(-4)}(n+1)+5040 \psi ^{(-3)}(n+1)-720 \psi ^{(-2)}(n+1)-10 \psi ^{(0)}(n+1)+\psi ^{(1)}(n+1))$$ $$=-810 \log (A)+3240 \zeta '(-7)+11340 \zeta '(-5)+7560 \zeta '(-3)+\frac{810 \zeta (3)}{\pi ^2}+\frac{42525 \zeta (7)}{\pi ^6}-\frac{8505 \zeta (5)}{\pi ^4}-\frac{127575 \zeta (9)}{2 \pi ^8}-\frac{\pi ^2}{6}+\frac{13371}{280}-9 \gamma +55 \log (2)-2 \log (32)+45 \log (\pi )$$

Y de nuevo, una última

$$\sum _{n=1}^{\infty }(380 \text{log$\Gamma $}(n+1)-2432902008176640000 \psi ^{(-19)}(n)+2432902008176640000 \psi ^{(-19)}(n+1)-2432902008176640000 \psi ^{(-18)}(n+1)+1216451004088320000 \psi ^{(-17)}(n+1)-405483668029440000 \psi ^{(-16)}(n+1)+101370917007360000 \psi ^{(-15)}(n+1)-20274183401472000 \psi ^{(-14)}(n+1)+3379030566912000 \psi ^{(-13)}(n+1)-482718652416000 \psi ^{(-12)}(n+1)+60339831552000 \psi ^{(-11)}(n+1)-6704425728000 \psi ^{(-10)}(n+1)+670442572800 \psi ^{(-9)}(n+1)-60949324800 \psi ^{(-8)}(n+1)+5079110400 \psi ^{(-7)}(n+1)-390700800 \psi ^{(-6)}(n+1)+27907200 \psi ^{(-5)}(n+1)-1860480 \psi ^{(-4)}(n+1)+116280 \psi ^{(-3)}(n+1)-6840 \psi ^{(-2)}(n+1)-20 \psi ^{(0)}(n+1)+\psi ^{(1)}(n+1))$$ $$=-7220 \log (A)+64980 \zeta '(-17)+1472880 \zeta '(-15)+10310160 \zeta '(-13)+28721160 \zeta '(-11)+35103640 \zeta '(-9)+19147440 \zeta '(-7)+4418640 \zeta '(-5)+368220 \zeta '(-3)+\frac{16245 \zeta (3)}{\pi ^2}+\frac{57994650 \zeta (7)}{\pi ^6}+\frac{62199262125 \zeta (11)}{\pi ^{10}}+\frac{11755660541625 \zeta (15)}{\pi ^{14}}+\frac{176334908124375 \zeta (19)}{2 \pi ^{18}}-\frac{1104660 \zeta (5)}{\pi ^4}-\frac{2261791350 \zeta (9)}{\pi ^8}-\frac{1119586718250 \zeta (13)}{\pi ^{12}}-\frac{58778302708125 \zeta (17)}{\pi ^{16}}-\frac{\pi ^2}{6}+\frac{17504273203}{16081065}-19 \gamma +210 \log (2)-2 \log (1024)+190 \log (\pi )$$ Ya decía yo que el anterior va a ser la última, pero esta es la última $$\sum_{n=1}^{\infty}(870 \text{log$\Gamma $}(n+1)-265252859812191058636308480000000 \psi ^{(-29)}(n)+265252859812191058636308480000000 \psi ^{(-29)}(n+1)-265252859812191058636308480000000 \psi ^{(-28)}(n+1)+132626429906095529318154240000000 \psi ^{(-27)}(n+1)-44208809968698509772718080000000 \psi ^{(-26)}(n+1)+11052202492174627443179520000000 \psi ^{(-25)}(n+1)-2210440498434925488635904000000 \psi ^{(-24)}(n+1)+368406749739154248105984000000 \psi ^{(-23)}(n+1)-52629535677022035443712000000 \psi ^{(-22)}(n+1)+6578691959627754430464000000 \psi ^{(-21)}(n+1)-730965773291972714496000000 \psi ^{(-20)}(n+1)+73096577329197271449600000 \psi ^{(-19)}(n+1)-6645143393563388313600000 \psi ^{(-18)}(n+1)+553761949463615692800000 \psi ^{(-17)}(n+1)-42597073035662745600000 \psi ^{(-16)}(n+1)+3042648073975910400000 \psi ^{(-15)}(n+1)-202843204931727360000 \psi ^{(-14)}(n+1)+12677700308232960000 \psi ^{(-13)}(n+1)-745747076954880000 \psi ^{(-12)}(n+1)+41430393164160000 \psi ^{(-11)}(n+1)-2180547008640000 \psi ^{(-10)}(n+1)+109027350432000 \psi ^{(-9)}(n+1)-5191778592000 \psi ^{(-8)}(n+1)+235989936000 \psi ^{(-7)}(n+1)-10260432000 \psi ^{(-6)}(n+1)+427518000 \psi ^{(-5)}(n+1)-17100720 \psi ^{(-4)}(n+1)+657720 \psi ^{(-3)}(n+1)-24360 \psi ^{(-2)}(n+1)-30 \psi ^{(0)}(n+1)+\psi ^{(1)}(n+1))$$ $$=-25230 \log (A)+353220 \zeta '(-27)+20663370 \zeta '(-25)+413267400 \zeta '(-23)+3734166150 \zeta '(-21)+17426108700 \zeta '(-19)+45149463450 \zeta '(-17)+67476121200 \zeta '(-15)+59041606050 \zeta '(-13)+30099642300 \zeta '(-11)+8713054350 \zeta '(-9)+1357878600 \zeta '(-7)+103316850 \zeta '(-5)+3178980 \zeta '(-3)+\frac{88305 \zeta (3)}{\pi ^2}+\frac{2324629125 \zeta (7)}{\pi ^6}+\frac{61753772705625 \zeta (11)}{2 \pi ^{10}}+\frac{179518217255251875 \zeta (15)}{\pi ^{14}}+\frac{735127099660256428125 \zeta (19)}{2 \pi ^{18}}+\frac{363887914331826931921875 \zeta (23)}{2 \pi ^{22}}+\frac{38208231004841827851796875 \zeta (27)}{4 \pi ^{26}}-\frac{30995055 \zeta (5)}{2 \pi ^4}-\frac{588131168625 \zeta (9)}{2 \pi ^8}-\frac{10559895132661875 \zeta (13)}{4 \pi ^{12}}-\frac{18849412811801446875 \zeta (17)}{2 \pi ^{16}}-\frac{40431990481314103546875 \zeta (21)}{4 \pi ^{20}}-\frac{7641646200968365570359375 \zeta (25)}{4 \pi ^{24}}-\frac{114624693014525483555390625 \zeta (29)}{8 \pi ^{28}}-\frac{\pi ^2}{6}-\frac{16571275403939220313}{40156716600}-29 \gamma +435 \log (2)+435 \log (\pi )$$

Answer 1

Para ahorrar algo de espacio, nos vamos a denotar $\psi_s(z+n)=:f^{(s)}(n)$. La serie en cuestión puede ser escrito como $$S=\sum_{n\ge1}\left(24f^{(-3)}(0)-24f^{(-3)}(1)+24f^{(-2)}(1)-12f^{(-1)}(1)+4f^{(0)}(1)-f^{(1)}(1)\right)$$ Observe que \begin{align} \int^1_0z^4f^{(2)}(n)\ {\rm d}z &=f^{(1)}(1)-\int^1_0 4z^3f^{(1)}(n)\ {\rm d}z\\ &=f^{(1)}(1)-4f^{(0)}(1)+\int^1_0 12z^2f^{(0)}(n)\ {\rm d}z\\ &=f^{(1)}(1)-4f^{(0)}(1)+12f^{(-1)}(1)-\int^1_0 24zf^{(-1)}(n)\ {\rm d}z\\ &=f^{(1)}(1)-4f^{(0)}(1)+12f^{(-1)}(1)-24f^{(-2)}(1)+\int^1_0 24f^{(-2)}(n)\ {\rm d}z\\ &=f^{(1)}(1)-4f^{(0)}(1)+12f^{(-1)}(1)-24f^{(-2)}(1)+24f^{(-3)}(1)-24f^{(-3)}(0) \end{align} Por lo $\displaystyle S=-\sum_{n\ge1}\int^1_0z^4\psi_2(z+n)\ {\rm d}z$. Desde \begin{align} -\sum_{n\ge1}\psi_2(z+n) =\int^1_0\frac{t^{z}\ln^2{t}}{(1-t)^2}{\rm d}t=-\int^1_0\frac{zt^{z-1}\ln^2{t}+2t^{z-1}\ln{t}}{1-t}{\rm d}t=z\psi_2(z)+2\psi_1(z) \end{align} Tenemos \begin{align} S =\int^1_0z^4\left(z\psi_2(z)+2\psi_1(z)\right)\ {\rm d}z =\frac{\pi^2}{6}-3\int^1_0z^4\psi_1(z)\ {\rm d}z =\frac{\pi^2}{6}+3\gamma+12\int^1_0z^3\psi_0(z)\ {\rm d}z \end{align} El resultado $$\color{red}{S=\frac{\pi^2}{6}+3\gamma-6\ln(2\pi)+3\ln{A}-\frac{9\zeta(3)}{\pi^2}}$$ se sigue inmediatamente de la fórmula $$\begin{align}\int_0^1 z^n\psi_0\left(z\right) {\rm d}z=&(-1)^{n-1}\zeta'(-n)+\frac{(-1)^nB_{n+1}H_n}{n+1}+\sum_{k=0}^n(-1)^k{n \choose k}\left[\zeta'(-k)-\frac{B_{k+1}H_k}{k+1}\right]\end{align}$$ demostrado en este documento, y el hecho de que $\displaystyle\zeta'(-2n)=\frac{(-1)^n\zeta(2n+1)(2n)!}{2^{2n+1}\pi^{2n}}$, lo que sigue a partir de Riemann funcional de la ecuación.

Uno debe ser capaz de acercarse a los otros de la serie en su pregunta con el mismo método.

Answer 2

No es trivial para mí que el término general es $o(1)$, pero en cualquier caso, me gustaría empezar a partir de la representación: $$ \psi(z) = \sum_{n\geq 0}\left(\log\left(1+\frac{1}{n+1}\right)-\frac{1}{n+z}\right) $$ y diferenciarla/ integrarlo termwise para demostrar que la suma es convergente y, finalmente, calcular su valor. Nunca antes había visto, de todos modos.