Htykuut

I would like to simplify the following expression,

$$sum_{n=0}^{infty}ne^{-a}frac{a^{x+n}}{(x+n)!}$$

where $x$ is an integer and $a<1$.

Is it possible to lose the sum?

An approximation for the sum will be also helpful.

The sum can be written as
$$
eqalign{
& S(a,x) = sumlimits_{n = 0}^infty {ne^{, - ,a} {{a^{,x + n} } over {left( {x + n} right)!}}} = cr
& = e^{, - ,a} sumlimits_{k = x}^infty {left( {k - x} right){{a^{,k} } over {k!}}} = e^{, - ,a} left( {sumlimits_{k = 0}^infty
{left( {k - x} right){{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 1} {left( {k - x} right){{a^{,k} } over {k!}}} } right) = cr
& = e^{, - ,a} left( {sumlimits_{k = 0}^infty {k{{a^{,k} } over {k!}}} - xsumlimits_{k = 0}^infty {{{a^{,k} } over {k!}}}
- sumlimits_{k = 0}^{x - 1} {k{{a^{,k} } over {k!}}} + xsumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } right) cr}
$$

Since
$$
left{ matrix{
e^{, - ,a} sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} = {{Gamma (x,a)} over {Gamma (x)}} = Q(x,a) hfill cr
sumlimits_{k = 0}^infty {k{{a^{,k} } over {k!}}} = a{d over {da}}sumlimits_{k = 0}^infty {{{a^{,k} } over {k!}}} = ae^{,,a} hfill cr
sumlimits_{k = 0}^{x - 1} {k{{a^{,k} } over {k!}}} = sumlimits_{k = 1}^{x - 1} {{{a^{,k} } over {left( {k - 1} right)!}}}
= asumlimits_{j = 0}^{x - 2} {{{a^{,j} } over {j!}}} = ae^{,,a} Q(x - 1,a) hfill cr} right.
$$

where $Q(x,a)$ is the Regularized Incomplete Gamma function,

it is easy to conclude.

Another way is
$$
eqalign{
& S(a,x) = sumlimits_{n = 0}^infty {ne^{, - ,a} {{a^{,x + n} } over {left( {x + n} right)!}}} = cr
& = e^{, - ,a} left( {sumlimits_{n = 0}^infty {left( {n + x} right){{a^{,x + n} } over {left( {x + n} right)!}}}
- xsumlimits_{n = 0}^infty {{{a^{,x + n} } over {left( {x + n} right)!}}} } right) = cr
& = e^{, - ,a} left( {asumlimits_{n = 0}^infty {{{a^{,x + n - 1} } over {left( {x + n - 1} right)!}}}
- xsumlimits_{n = 0}^infty {{{a^{,x + n} } over {left( {x + n} right)!}}} } right) = cr
& = e^{, - ,a} left( {asumlimits_{n = 0}^infty {{{a^{,left( {x - 1} right) + n} } over {left( {left( {x - 1} right) + n} right)!}}}
- xsumlimits_{n = 0}^infty {{{a^{,left( {x - 1} right) + n + 1} } over {left( {left( {x - 1} right) + n + 1} right)!}}} } right) = cr
& = e^{, - ,a} left( {asumlimits_{n = 0}^infty {{{a^{,left( {x - 1} right) + n} } over {left( {left( {x - 1} right) + n} right)!}}}
- xsumlimits_{n = 1}^infty {{{a^{,left( {x - 1} right) + n} } over {left( {left( {x - 1} right) + n} right)!}}} } right) = cr
& = e^{, - ,a} left( {a{{a^{,left( {x - 1} right)} } over {left( {x - 1} right)!}} + asumlimits_{n = 1}^infty {{{a^{,left( {x - 1} right) + n} }
over {left( {left( {x - 1} right) + n} right)!}}}
- xsumlimits_{n = 1}^infty {{{a^{,left( {x - 1} right) + n} } over {left( {left( {x - 1} right) + n} right)!}}} } right) = cr
& = e^{, - ,a} left( {a{{a^{,left( {x - 1} right)} } over {left( {x - 1} right)!}}
+ left( {a - x} right)sumlimits_{n = 0}^infty {{{a^{,x + n} } over {left( {x + n} right)!}}} } right) = cr
& = e^{, - ,a} left( {a{{a^{,left( {x - 1} right)} } over {left( {x - 1} right)!}}
+ left( {a - x} right)sumlimits_{k = x}^infty {{{a^{,k} } over {k!}}} } right) = cr
& = e^{, - ,a} left( {a{{a^{,left( {x - 1} right)} } over {left( {x - 1} right)!}} + left( {a - x} right)sumlimits_{k = 0}^infty {{{a^{,k} } over {k!}}}
- left( {a - x} right)sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } right) = cr
& = e^{, - ,a} {{a^{,x} } over {left( {x - 1} right)!}} + left( {a - x} right)
- left( {a - x} right)e^{, - ,a} sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} = cr
& = e^{, - ,a} {{a^{,x} } over {Gamma (x)}} + left( {a - x} right)left( {1 - Q(x,a)} right) = cr
& = e^{, - ,a} {{a^{,x} } over {Gamma (x)}} + left( {a - x} right){{gamma (x,a)} over {Gamma (x)}} cr}
$$
which checks with the previous one.

--- Addendum ---

Concerning your comment and the request for a nice approximation to the sum $S(a,x)$,
it is not much clear what you intend / need to do. In any case consider the following:

- the formula above (1st and 2nd way lead to the same result) is valid for $a$ and $c$
real or even complex ( with some limitations);

- if $x$ is an integer as you say, then for small values of it the expression in the last-but-third line
is easily computable;

- if $x$ is large , we can apply to $Gamma(x)$ and $Q(x,a)$ the known asymptotic expansions
which are
$$
eqalign{
& Gamma (x) = sqrt {,{{2,pi } over x},} left( {{x over e}} right)^{,x} left( {1 + Oleft( {{1 over x}} right)} right)
quad left| {;x, to ,infty ,;;left| {,arg (x),} right|} right. < pi cr
& Q(x,a) approx 1 - {{x^{, - x - 1/2} e^{,x - a} a^{,x} } over {sqrt {2pi } }}left( {1 + {{12a - 1} over {12x}}
+ Oleft( {{1 over {x^{,2} }}} right)} right);quad left| {,left| x right|; to ;infty } right. cr
& Q(x,a) approx Q(x,1) - {{e^{, - 1} } over {Gamma (x)}}left( {a - 1} right) - {{e^{, - 1} left( {x - 2} right)} over {2Gamma (x)}}
left( {a - 1} right)^{,2} + Oleft( {left( {a - 1} right)^{,3} } right);quad left| {,left| a right|; to ;1} right. cr}
$$

$$sum_{n=0}^infty ne^{-a}frac{a^{x+n}}{(x+n)!}=sum_{n=0}^infty frac{a^x}{e^a}frac{na^n}{(x+n)!}=frac{a^x}{e^a}sum_{n=0}^infty a^nfrac{n}{(x+n)!};$$
$$scriptstylesum_{n=0}^infty a^nfrac{n}{(x+n)!}=e^aa^{1 - x} - frac{e^aa^{1 - x}Gamma(x + 1, a)}{Γ(x + 1)}- e^axa^{-x}+frac{e^aa^{-x}Gamma(x + 1, a)}{Gamma(x)} + frac{a}{Gamma(x + 1)}=e^aa^{1-x}-frac{e^aa^{1-x}}{x!}frac{x!}{e^a}sum_{k=0}^{x}frac{a^k}{k!}-frac{e^ax}{a^x}+frac{e^a}{a^xx!}sum_{k=0}^{x}frac{a^k}{k!}+frac{a}{x!}=e^aa^{1-x}-a^{1-x}sum_{k=0}^{x}frac{a^k}{k!}-frac{e^ax}{a^x}+frac{e^a}{a^xx!}sum_{k=0}^{x}frac{a^k}{k!}+frac{a}{x!}=e^aa^{1-x}+left(frac{e^a}{a^xx!}-a^{1-x}right)sum_{k=0}^xfrac{a^k}{k!}-frac{e^ax}{a^x}+frac{a}{x!};$$
$$scriptstylefrac{a^x}{e^a}sum_{n=0}^infty a^nfrac{n}{(x+n)!}=frac{a^x}{e^a}left(e^aa^{1-x}+left(frac{e^a}{a^xx!}-a^{1-x}right)sum_{k=0}^xfrac{a^k}{k!}-frac{e^ax}{a^x}+frac{a}{x!}right)$$
$$=a+left(frac{1}{x!}-frac{a}{e^a}right)sum_{k=0}^xfrac{a^k}{k!}-x+frac{a^{x+1}}{e^ax!}$$
That's the best I can do, it's a lot of algebra, so you might want to double check, but I think that's as simplified as it can possibley get.

This is not an algebraic function, so it can only really be written in terms of improper integrals or appropriate special functions.

https://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D0%7D%5E%5Cinfty+ne%5E%7B-a%7D%5Cfrac%7Ba%5E%7Bx%2Bn%7D%7D%7B(x%2Bn)!%7D

Starting from R. Burton's answer, the result could simplify to
$$sum_{n=0}^{infty}ne^{-a}frac{a^{x+n}}{(x+n)!}=frac{a^{x+1} }{x!}left((x-a) E_{-x}(a)+e^{-a}right)+a-x$$ where appears the exponential integral function.