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 $e=2.718281828459...$

  • $\begingroup$ My quick suggestion from a moving train would be to explore the properties of convolution sums. $\endgroup$ Commented Jun 2, 2017 at 11:11

1 Answer 1


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.

  • 2
    $\begingroup$ In the denominators of the limits ? ... $1-x(exp(\color{red}{1}-x))$ ... ? $\endgroup$ Commented Jun 2, 2017 at 19:32
  • 1
    $\begingroup$ Sorry, small typo. It's fixed now. $\endgroup$ Commented Jun 2, 2017 at 19:47
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    $\begingroup$ Maybe another typo in the first sum ? $n \sum_{k=0}^n \frac{(-1)^{n-k} k^{n-k\color{red}{-1}}}{(n-k)!} e^k$ $\endgroup$ Commented Jun 3, 2017 at 11:04
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    $\begingroup$ Well spotted. I've fixed it. $\endgroup$ Commented Jun 3, 2017 at 12:57

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