# How to compute $(-1)^{n+1}n!(1-e\sum_{k=0}^n\frac{(-1)^k}{k!})$?

I was doing some work on the integral $$\int_0^1 x^ne^x dx$$ and I eventually came to this expression in terms of n $$\int_0^1 x^ne^x dx=(-1)^{n+1}n!\biggl(1-e\sum_{k=0}^n\frac{(-1)^k}{k!}\biggr)$$ Now my question is this, how do I compute the limit of this expression as $$n$$ goes to $$\infty$$? I recognize that $$\lim_{n\to\infty}e\sum_{k=0}^{n}\frac{(-1)^k}{k!}=1$$ per the series expression of $$e^x$$ at $$-1$$. This gives me an indeterminate form, but I can't apply L'Hopital's rule because I can't differentiate $$n!$$ (or $$1/n!$$ as the case would be after re-arranging), and I also can't think of a variable substitution that would help me here. Common sense and wolfram definitely imply that the limit is in fact $$0$$, but how can I show this?

• $n!=\Gamma(n+1)$ is differentiable. – J.G. Nov 15 at 7:01
• perhaps some properties of the gamma function/complex analysis would be useful – rubikscube09 Nov 15 at 7:04
• I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval? – JacksonFitzsimmons Nov 15 at 7:06

When $$x\in [0,1]$$, $$e^x\le e$$, by first mean value theorem for definite integrals, \begin{align*} I_n&\le\max_{x\in[0,1]}\{e^x\}\int_0^1x^{n}~\mathrm dx\\ &=\frac{e}{n+1}\to0. \end{align*} Since $$I_n>0$$, the limit is $$0$$ by squeeze theorem.
Not as slick as Tianlalu's method, but for $$y\in (0,\,1)$$ we have $$\lim_{n\to\infty}\int_0^y x^n e^x dx \le \lim_{n\to\infty}y^n\int_0^y e^x dx=0.$$
$$\exp(-1)=\sum_{k=0}^n\frac{(-1)^k}{k!}+\frac{e^\xi}{(n+1)!}(-1)^{n+1}$$
for some $$-1<\xi<0$$. If you estimate $$e^\xi$$ and replace the sum you get convergence to $$0$$.