Please help me to calculate the sum $t \sum_{s=0}^{t} \frac{ (-1)^{t-s} s^{t-s-1} e^s }{(t - s)!}$ I need to prove that the following sum
$$t \sum_{s=0}^{t} \frac{ (-1)^{t-s} s^{t-s-1} e^s }{(t - s)!}$$
tends to 2, when t goes to infinity.
Here $e=2.718281828459...$
 A: I suspect there will be some trick to make things easier, but I haven't found it yet. I did find an answer, although it might not be the most straightforward.
Anyway, let's rewrite the sum as follows:
$$
\begin{align}
n \sum_{k=0}^n \frac{(-1)^{n-k} k^{n-k-1}}{(n-k)!} e^k
&= n \sum_{t+s=n} \frac{(-1)^t s^{t-1}}{t!} e^s\\
&= \sum_{t+s=n} (t+s) \frac{(-1)^t s^{t-1}}{t!} e^s\\
&= \sum_{t+s=n} \frac{(-1)^t s^{t}}{t!} e^s  - \frac{(-1)^{t-1} s^{t-1}}{(t-1)!} e^s\\
&= \sum_{t+s=n} \frac{(-s)^t}{t!} e^s  - \frac{(-s)^{t-1}}{(t-1)!} e^s\\
\end{align}
$$
(for the sake of argument we'll let $1 / (-1)!  = 0$)
Now two rather curious things have happened, first of all it seems to have turned into a not quite telescoping sum (which doesn't turn out to be too useful), but more importantly we can also recognise $\frac{(-s)^t}{t!}$, which is just begging to be turned into a series for $\exp(-s)$. Unfortunately $t$ and $s$ depend on one another, which makes things rather difficult. To avoid this consider instead the series:
$$
\begin{align}
&\sum_{n=0}^\infty \left(\sum_{t+s=n} \frac{(-s)^t}{t!} e^s  - \frac{(-s)^{t-1}}{(t-1)!} e^s \right)x^{n} \\
&= \sum_{s=0}^\infty \sum_{t=0}^\infty \left(\frac{(-s)^t}{t!} e^s  - \frac{(-s)^{t-1}}{(t-1)!} e^s \right)x^{s+t}\\
&= \sum_{s=0}^\infty \sum_{t=0}^\infty \frac{(-xs)^t}{t!} (xe)^s  - x \frac{(-xs)^{t-1}}{(t-1)!} (xe)^s \\
&= \sum_{s=0}^\infty  \exp(-xs) (xe)^s  - x \exp(-xs)(xe)^s\\
&= \sum_{s=0}^\infty (1 - x)(xe^{1-x})^s\\
&= \frac{1-x}{1 - xe^{1-x}}
\end{align}
$$
Let's denote this $f(x)$. Now to show that the sum we were interested in converges to 2, we'd need to show that these series coefficients converge to 2, which is equivalent to saying that $f(x) = \frac2{1-x} + h(x)$ for some analytic function $h(x)$. The easiest way of doing this is to use l'Hopitals rule to show:
$$
\lim_{x\to1} \frac{(1-x)^2 }{ 1 - x (\exp(1-x))}
= \lim_{x\to1} \frac{-2(1-x)}{x \exp(1-x) - \exp(1-x)}
= \lim_{x\to1} \frac{2}{\exp(1-x)} = 2
$$
Since $f(x)$ is analytic everywhere else we can write $(1-x)f(x) = 2 + (1-x) h(x)$ for some analytic function $h$, therefore $f = \frac2{1-x} + h(x)$ so we're done.
The alternative method is to calculate $\frac{(1-x)^2 }{ 1 - x (\exp(1-x))} - \frac{2}{1-x}$ and show that it converges, which also works, but is quite a bit more work.
