Forma corta de algunas series

Question

Hay un breve formulario para la suma de la siguiente serie?

$$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}((2y-1)^{2k+1}+1)}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$$ $$\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}(\cos^{-1}(2y-1)-\pi)}{2^{4n+3}n!(n+1)!}$$ $$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n-2k+3}n!(n+1)!(2k+1)!}$$

Incluso la suma de cualquier combinación de los anteriores términos podría ayudar. Este es el resultado de algunos integral. Supongo que debe contener Bessel y funciones de Struve. En el caso de $y=0.5$ parece que la suma de los términos arriba mencionados, para ser

$$ -\frac{\pi}{2}\left[I_3(\frac{\alpha}{2}) +\frac{3}{\frac{\alpha}{2}}I_2(\frac{\alpha}{2})- I_{1}(\frac{\alpha}{2})-\frac{1}{8}\left[L_{-3}(\frac{\alpha}{2}) - L_{-1}(\frac{\alpha}{2}) - L_{1}(\frac{\alpha}{2})+L_{3}(\frac{\alpha}{2}) -\frac{2\alpha^{-2}}{\pi} -\frac{8}{3\pi}+\frac{2\alpha^2}{15\pi}\right]\right] $$

Answer 1

Para la primera serie, pude wittle hacia abajo hasta una suma de una función de Struve y una serie no evaluada de funciones hipergeométrica:

$$\begin{align} S{(\alpha,y)} &=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}((2y-1)^{2k+1}+1)}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}\\ &=\sum\limits_{n=0}^\infty \frac{\alpha^{2n}}{2^{2n+1}(2n)!} \sum\limits_{k=0}^n (-1)^k \frac{n!}{k!(n-k)!} \dfrac{((2y-1)^{2k+1}+1)}{(2k+1)}\\ &=\sum\limits_{n=0}^\infty \frac{\alpha^{2n}}{2^{2n+1}(2n)!} \left[\sum\limits_{k=0}^n \dfrac{(-1)^k\binom{n}{k}}{(2k+1)} + \sum\limits_{k=0}^n \dfrac{(-1)^k\binom{n}{k}}{(2k+1)} (2y-1)^{2k+1}\right]\\ &=\sum\limits_{n=0}^\infty \frac{\alpha^{2n}}{2^{2n+1}(2n)!} \left[\frac{(2n)!!}{(2n+1)!!} + (2y-1)\,{_2F_1}{\left(\frac12,-n;\frac32;(2y-1)^2\right)}\right]\\ &=\sum\limits_{n=0}^\infty \frac{\alpha^{2n}}{2^{2n+1}(2n)!}\frac{(2n)!!}{(2n+1)!!} + (2y-1)\sum\limits_{n=0}^\infty \frac{\alpha^{2n}}{2^{2n+1}(2n)!} {_2F_1}{\left(\frac12,-n;\frac32;(2y-1)^2\right)}\\ &=\frac{\pi}{4}\operatorname{L}_{-1}{\left(\frac{\alpha}{2}\right)} + (2y-1)\sum\limits_{n=0}^\infty \frac{\alpha^{2n}}{2^{2n+1}(2n)!} {_2F_1}{\left(\frac12,-n;\frac32;(2y-1)^2\right)}. \end {Alinee el} $$

Answer 2

Para $\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}(2y-1)^{2k+1}}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$ ,

$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}(2y-1)^{2k+1}}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^kn!\alpha^{2n}(2y-1)^{2k+1}}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^k(n+k)!\alpha^{2n+2k}(2y-1)^{2k+1}}{2^{2n+2k+1}(2n+2k)!k!n!(2k+1)}$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^k\alpha^{2n+2k}(2y-1)^{2k+1}\sqrt\pi}{2^{4n+4k+1}\Gamma\left(n+k+\dfrac{1}{2}\right)n!k!\left(k+\dfrac{1}{2}\right)}$ (de acuerdo a https://en.wikipedia.org/wiki/Multiplication_theorem#Gamma_function-Legendre_function)

Para $\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n-2k+3}n!(n+1)!(2k+1)!}$ ,

$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n-2k+3}n!(n+1)!(2k+1)!}$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(k!)^2\alpha^{2n+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n-2k+3}n!(n+1)!(2k+1)!}$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(k!)^2\alpha^{2n+2k+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n+2k+3}(n+k)!(n+k+1)!(2k+1)!}$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(k!)^2\alpha^{2n+2k+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}\sqrt\pi}{16^{n+k+1}(n+k)!(n+k+1)!\Gamma(k+1)\Gamma\left(k+\dfrac{3}{2}\right)}$ (de acuerdo a https://en.wikipedia.org/wiki/Multiplication_theorem#Gamma_function-Legendre_function)

que ambos se refieren a Kampé de Fériet función función