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Is there an easy closed-form term for

$$\sum_{j=k}^{\infty} \frac{x^j}{j!}e^{-x},$$

thus when the sum starts at a constant $k$ instead of $1$?

EDIT: Thanks for your help. Is there a Chance to solve this sum-term? Because this is not really what I expect, when I talk about a closed-form term.

A Little bit more of context might help, maybe:

I have $$f(n,p)=\sum_{j=k}^{\infty} \frac{(np)^j}{j!} e^{-np}$$ and it is meant that the partial Derivation is $$\frac{\delta f(n,p)}{\delta n}=\frac{p (np)^{k-1}}{(k-1)!}e^{-np}$$ but I have no idea how to get to this.

Because to me:

$$\frac{\delta f(n,p)}{\delta n}=\sum_{j=k}^{\infty} \left( \frac{p (np)^{j-1}}{j!} e^{-np} -\frac{p (np)^j}{j!} e^{-np} \right)$$ but then I am stuck.

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1  
See CDF of Poisson distribution. –  Tunk-Fey Jul 7 at 12:49
1  
From $\displaystyle{\large\tt M}$: $\displaystyle{\large\frac{\Gamma (k)-\Gamma (k,x)}{\Gamma (k)}}$. –  Felix Marin Jul 7 at 17:03
    
@FelixMarin Could you specify what you mean with that? I know about the Gamma and the incomplete Gamma functions, but I can't see how to apply this here. –  user136457 Jul 7 at 17:05
    
I just put your expression in $\tt Mathematica$. –  Felix Marin Jul 7 at 18:53

4 Answers 4

up vote 0 down vote accepted

You just made a mistake when differentiating $$f(x)=\sum_{j=k}^{\infty} \frac{(xp)^j}{j!} e^{-xp}.$$ The actual derivative is $$f'(x)=\sum_{j=k}^{\infty} \left( \color{red}{j}\,\frac{p\,(xp)^{j-1}}{j!} e^{-xp} -\frac{p\,(xp)^j}{j!} e^{-xp} \right),$$ that is, provided $k\geqslant1$, using the change of indices $\ell=j-1$ in the first summation, $$f'(x)=\sum_{\ell=k-1}^{\infty}\frac{p(xp)^{\ell}}{\ell!} e^{-xp} -\sum_{j=k}^{\infty} \frac{p (xp)^j}{j!} e^{-xp}=\frac{p(xp)^{k-1}}{(k-1)!} e^{-xp}.$$

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$$ \sum_{j=k}^{\infty} \frac{x^j}{j!}e^{-x}= \left(\sum_{j=k}^{\infty} \frac{x^j}{j!}+\sum_{j=0}^{k-1} \frac{x^j}{j!}-\sum_{j=0}^{k-1} \frac{x^j}{j!}\right)e^{-x}\\ =\left(e^x-\sum_{j=0}^{k-1} \frac{x^j}{j!}\right)e^{-x}=1-e^{-x}\sum_{j=0}^{k-1} \frac{x^j}{j!} $$

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thanks for your answer. I stated above in the question the whole Problem, pointing out what I could mean with a closed form, because somehow I don't get how they got to this solution. –  user136457 Jul 7 at 12:49
    
@mookid correct but my sum's upper limit is $k-1$, or did I miss something... –  draks ... Jul 7 at 14:34
    
you are right. But I don't know if it is what OP was looking for.... –  mookid Jul 7 at 14:44

Yes:

$$e^{-x}\sum_{j=k}^\infty \frac 1{j!} x^j = e^{-x}\left[\exp x - \sum_{j=0}^{k-1} \frac 1{j!} x^j \right] =e^{-x}\int_0^x \frac{(x-t)^{k-1}}{(k-1)!} e^{t} dt $$

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1  
$x$ as limit and inside the integral looks a little strange. Are you sure? –  draks ... Jul 7 at 12:50
    
x does not appear as a limit, does it? –  mookid Jul 7 at 13:02
    
$e^{-x}\int_0^{\color{red}x}$... –  draks ... Jul 7 at 14:32
    
there is no error, see en.wikipedia.org/wiki/… –  mookid Jul 7 at 14:43
    
@draks...: it would only seem strange to me if $x$ were the variable of integration, but $t$ is the variable of integration. –  robjohn Jul 8 at 15:51

Incomplete Gamma function and Eq.2 in : http://mathworld.wolfram.com/IncompleteGammaFunction.html

enter image description here

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