How can I evaluate $sum_m=0^infty left(frac(m+n)!n!right)^a frac(-x)^mm!$?

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I have the infinite sum



$$sum_m=0^infty left(frac(m+n)!n!right)^a frac(-x)^mm!$$



with $ain [0,1]$ and $nin mathbbN$ and try to calculate a closed form for it. For $a=0$ this is just the Taylor expansion of $e^-x$ and for $a=1$ it is the Taylor expansion of $(x+1)^-(n+1)$.



I already tried to interpret the sum as a Taylor expansion and was looking for a generalization of $(x+1)^-(n+1)$, but without success.



If there is a way, to show at least that the result is positive within the radius of convergence, it would be helpful to.






sequences-and-series taylor-expansion







share|cite|improve this question



















  • 1




    Question; Did you actually mean "for $a=0$ it is the Taylor expansion of $(x+1)^-(n+1)$" or " "for $a=1$ it is the Taylor expansion of $(x+1)^-(n+1)$"?
    – MathOverview
    Sep 10 at 13:46











  • Thanks, I fixed this mistake.
    – TheIdealis
    Sep 10 at 14:17






  • 1




    No hope for a closed-form for the $a$ in the given range except for $a=0,1.$ For $a=-1,$ you have a Bessel function. Asymptotics as $x to infty$ might be interesting.
    – skbmoore
    Sep 10 at 16:37











  • Right, I haven't noticed the relation to the Bessel function yet. The problem is that the series only converges for $x<1$ (for $a > 0$).
    – TheIdealis
    Sep 11 at 15:26














up vote
3
down vote

favorite












I have the infinite sum



$$sum_m=0^infty left(frac(m+n)!n!right)^a frac(-x)^mm!$$



with $ain [0,1]$ and $nin mathbbN$ and try to calculate a closed form for it. For $a=0$ this is just the Taylor expansion of $e^-x$ and for $a=1$ it is the Taylor expansion of $(x+1)^-(n+1)$.



I already tried to interpret the sum as a Taylor expansion and was looking for a generalization of $(x+1)^-(n+1)$, but without success.



If there is a way, to show at least that the result is positive within the radius of convergence, it would be helpful to.






sequences-and-series taylor-expansion







share|cite|improve this question



















  • 1




    Question; Did you actually mean "for $a=0$ it is the Taylor expansion of $(x+1)^-(n+1)$" or " "for $a=1$ it is the Taylor expansion of $(x+1)^-(n+1)$"?
    – MathOverview
    Sep 10 at 13:46











  • Thanks, I fixed this mistake.
    – TheIdealis
    Sep 10 at 14:17






  • 1




    No hope for a closed-form for the $a$ in the given range except for $a=0,1.$ For $a=-1,$ you have a Bessel function. Asymptotics as $x to infty$ might be interesting.
    – skbmoore
    Sep 10 at 16:37











  • Right, I haven't noticed the relation to the Bessel function yet. The problem is that the series only converges for $x<1$ (for $a > 0$).
    – TheIdealis
    Sep 11 at 15:26












up vote
3
down vote

favorite









up vote
3
down vote

favorite











I have the infinite sum



$$sum_m=0^infty left(frac(m+n)!n!right)^a frac(-x)^mm!$$



with $ain [0,1]$ and $nin mathbbN$ and try to calculate a closed form for it. For $a=0$ this is just the Taylor expansion of $e^-x$ and for $a=1$ it is the Taylor expansion of $(x+1)^-(n+1)$.



I already tried to interpret the sum as a Taylor expansion and was looking for a generalization of $(x+1)^-(n+1)$, but without success.



If there is a way, to show at least that the result is positive within the radius of convergence, it would be helpful to.






sequences-and-series taylor-expansion







share|cite|improve this question















I have the infinite sum



$$sum_m=0^infty left(frac(m+n)!n!right)^a frac(-x)^mm!$$



with $ain [0,1]$ and $nin mathbbN$ and try to calculate a closed form for it. For $a=0$ this is just the Taylor expansion of $e^-x$ and for $a=1$ it is the Taylor expansion of $(x+1)^-(n+1)$.



I already tried to interpret the sum as a Taylor expansion and was looking for a generalization of $(x+1)^-(n+1)$, but without success.



If there is a way, to show at least that the result is positive within the radius of convergence, it would be helpful to.






sequences-and-series taylor-expansion




sequences-and-series taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 10 at 14:17

























asked Sep 10 at 13:41









TheIdealis

162




162







  • 1




    Question; Did you actually mean "for $a=0$ it is the Taylor expansion of $(x+1)^-(n+1)$" or " "for $a=1$ it is the Taylor expansion of $(x+1)^-(n+1)$"?
    – MathOverview
    Sep 10 at 13:46











  • Thanks, I fixed this mistake.
    – TheIdealis
    Sep 10 at 14:17






  • 1




    No hope for a closed-form for the $a$ in the given range except for $a=0,1.$ For $a=-1,$ you have a Bessel function. Asymptotics as $x to infty$ might be interesting.
    – skbmoore
    Sep 10 at 16:37











  • Right, I haven't noticed the relation to the Bessel function yet. The problem is that the series only converges for $x<1$ (for $a > 0$).
    – TheIdealis
    Sep 11 at 15:26












  • 1




    Question; Did you actually mean "for $a=0$ it is the Taylor expansion of $(x+1)^-(n+1)$" or " "for $a=1$ it is the Taylor expansion of $(x+1)^-(n+1)$"?
    – MathOverview
    Sep 10 at 13:46











  • Thanks, I fixed this mistake.
    – TheIdealis
    Sep 10 at 14:17






  • 1




    No hope for a closed-form for the $a$ in the given range except for $a=0,1.$ For $a=-1,$ you have a Bessel function. Asymptotics as $x to infty$ might be interesting.
    – skbmoore
    Sep 10 at 16:37











  • Right, I haven't noticed the relation to the Bessel function yet. The problem is that the series only converges for $x<1$ (for $a > 0$).
    – TheIdealis
    Sep 11 at 15:26







1




1




Question; Did you actually mean "for $a=0$ it is the Taylor expansion of $(x+1)^-(n+1)$" or " "for $a=1$ it is the Taylor expansion of $(x+1)^-(n+1)$"?
– MathOverview
Sep 10 at 13:46





Question; Did you actually mean "for $a=0$ it is the Taylor expansion of $(x+1)^-(n+1)$" or " "for $a=1$ it is the Taylor expansion of $(x+1)^-(n+1)$"?
– MathOverview
Sep 10 at 13:46













Thanks, I fixed this mistake.
– TheIdealis
Sep 10 at 14:17




Thanks, I fixed this mistake.
– TheIdealis
Sep 10 at 14:17




1




1




No hope for a closed-form for the $a$ in the given range except for $a=0,1.$ For $a=-1,$ you have a Bessel function. Asymptotics as $x to infty$ might be interesting.
– skbmoore
Sep 10 at 16:37





No hope for a closed-form for the $a$ in the given range except for $a=0,1.$ For $a=-1,$ you have a Bessel function. Asymptotics as $x to infty$ might be interesting.
– skbmoore
Sep 10 at 16:37













Right, I haven't noticed the relation to the Bessel function yet. The problem is that the series only converges for $x<1$ (for $a > 0$).
– TheIdealis
Sep 11 at 15:26




Right, I haven't noticed the relation to the Bessel function yet. The problem is that the series only converges for $x<1$ (for $a > 0$).
– TheIdealis
Sep 11 at 15:26















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