High school contest question Some work on it reveals the possibility of using gamma function. Is there any easy way to compute it?
$$\lim_{n\to\infty}\left(\frac{1}{n!} \int_0^e \log^n x \ dx\right)^n$$
 A: Let $ x = e^{-y} $ to arrive at $ \displaystyle(-1)^n \int_{-1}^\infty  y^n e^{-y} \ dy = (-1)^n \left(n! + \int_{-1}^0 y^n e^{-y} \ dy\right) $. This second integral is bounded between $ 0 $ and $ (-1)^n $. In fact, it is exactly equal to $ u_n e - n! $ where $ u_n $ has a recursive definition easily found though integration by parts, namely $ u_{n} = (-1)^{n} + nu_{n-1} $.
Hence, the overall integral has value $ (-1)^n u_n e $. Because $ u_n \sim \frac{n!}{e} $ the overall result is $ (-1)^{n^2} = (-1)^n  $. 
The latter result can be drawn from the fact that $\displaystyle \lim_{n \to \infty} \int_{-1}^0 y^n e^{-y} \ dy = \lim_{n \to \infty} u_n e - n! = 0 $. 
A more interesting question is my opinion is $$ \sum_{n = 1}^\infty \left(\frac{1}{n!}\int_0^e \log^n x \ dx\right)^n - (-1)^n $$
A: I'm not totally sure of this, but Mathematica seemed to agree. Via integration by parts, $u=\log^{n}(x)\implies du = (n/x)\log^{n-1}(x)$ and $dv = dx\implies v = x$. Therefore,:
$$
\int_{0}^{e}\log^{n}(x)dx = e - n\int_{0}^{e}\log^{n-1}(x)
$$
By defining $a_{n} = \int_{0}^{e}\log^{n}(x)dx$, we see that $a_{n} = e - n a_{n-1}$. If $n$ is very large, the term iwth $e$ becomes unimportant, and effectively $a_{n} = -na_{n-1}$, which means that $a_{n} \sim (-1)^n n!$. Subbing this in, the limit $L$ becomes
$$
L\sim\lim_{n\to\infty}\left(\frac{(-1)^n n!}{n!}\right)^n = \lim_{n\to\infty}\left((-1)^n\right)^n = \lim_{n\to\infty}(-1)^{n^2}
$$
Which oscillates between $-1$ and $1$ for odd and even $n$, so the limit does not exist.
