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

Question

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?

Answer 1

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.

Answer 2

I don't know what you are 'allowed' to use but you could use the lagrange remainer for the exponential function:

$$\exp(-1)=\sum_{k=0}^n\frac{(-1)^k}{k!}+\frac{e^\xi}{(n+1)!}(-1)^{n+1}$$

for some $-1<\xi<0$. If you estimate $e^\xi $ and replace the sum you get convergence to $0$.

Answer 3

Not as slick as Tianlalu's method, but for $y\in (0,\,1)$ we have $$ \lim_{n\to\infty}\int_0^y x^n e^x dx \le \lim_{n\to\infty}y^n\int_0^y e^x dx=0. $$