$\sum_{k=n}^{\infty}\left(n-k\right)e^{-\lambda}\frac{\lambda^{k}}{k!}= ?$ Could you please help me. How do I sum the following: 
$$\sum_{k=n}^{\infty}\left(n-k\right)e^{-\lambda}\frac{\lambda^{k}}{k!}$$
If the summation had started at 0, then it would be simply an expectation of (n-k), where K is the Poisson distribution.
 A: Taylor series:
$$ e^{\lambda x} = \sum^{\infty}_{k=0} \frac {(\lambda x) ^k}{k!}$$
We get
$$e^{\lambda } = \sum^{\infty}_{k=0} \frac {\lambda  ^k}{k!} $$
Simplify the equation in the question:
$$\sum_{k=n}^{\infty}\left(n-k\right)e^{-\lambda}\frac{\lambda^{k}}{k!} = n e^{-\lambda}\sum_{k=n}^{\infty}\frac{\lambda^{k}}{k!}-\lambda e^{-\lambda}\sum_{k=n}^{\infty}\frac{\lambda^{k-1}}{(k-1)!} =n-\lambda-ne^{-\lambda}\sum^{n-1}_{k=0} \frac {\lambda  ^k}{k!}+\lambda e^{-\lambda}\sum^{n-2}_{k=0} \frac {\lambda  ^k}{k!}$$
Hence the result is:
$$(n-\lambda)(1-e^{-\lambda}\sum^{n-2}_{k=0} \frac {\lambda  ^k}{k!})-ne^{-\lambda}\frac {\lambda  ^{n-1}}{(n-1)!}$$
where $n\geq 2$
Special cases:
1.$n=0$ , $-\lambda$
2.$n=1$ , $1-\lambda-e^{-\lambda}$
A: If you set
$$
I_k(\lambda):=\int_0^{\lambda}x^ke^{-x}\:dx, \qquad k=0,1,2\cdots.\tag1
$$ Then, integrating by parts, you have
$$
\begin{align}I_{k+1}(\lambda)&=\left[x^{k+1}\left(-e^{-x}\right)\right]_0^{\lambda}+(k+1)\int_0^{\lambda}x^ke^{-x}\:dx
\end{align}
$$or$$
I_{k+1}(\lambda)=-\lambda^{k+1}e^{-\lambda}+(k+1)I_k(\lambda), \quad k\geq1 \tag2.
$$ From $(2)$, you deduce
$$
\sum _{k=n}^{\infty} e^{-\lambda }\frac{\lambda ^k}{k!}=\frac{1}{(n-1)!}\int _0^{\lambda }x^{n-1}e^{-x}dx \quad n=1,2,\cdots,\tag3
$$ or, using the incomplete gamma function, you get

$$
(n-k)\sum _{k=n}^{\infty} e^{-\lambda }\frac{\lambda ^k}{k!}=(n-k)-\frac{(n-k)}{(n-1)!}\gamma(n,\lambda) \qquad n=1,2,\cdots.\tag4
$$ 

From $(4)$, you may obtain the asymptotics,
$$
(n-k)\left(1-e^{-\lambda}\frac{\lambda^n}{n!} \left(\frac{1}{\lambda}+\frac{n-1}{\lambda ^2}+O\left(\frac1{\lambda}\right)^3\right)\right),\quad \lambda \to \infty,
$$ or
$$
(n-k)\left(1+\frac{\lambda^n}{n!} e^{-\lambda }\left(\frac1{n}+\frac1{\lambda }{n^2}+O\left(\frac{1}{n}\right)^3\right)\right),\quad n \to \infty.
$$
