# High school contest question

Some work on it reveals the possibility of using gamma function. Is there any easy way to compute it? $$\lim_{n\to\infty}\left(\frac{1}{n!} \int_0^e \log^n x \ dx\right)^n$$

-
"Some work..." what exactly did you do that makes you expect the gamma function would be useful? –  Stahl May 18 '13 at 15:49
@Stahl I let $x=e^{-y}$. Since it's a high school question I expect to find a way that avoids the use of the incomplete gamma function. –  Student_kid May 18 '13 at 15:51
That's fair - I only ask because it's easier to give suitable answers when one exhibits his or her previous work on the problem. –  Stahl May 18 '13 at 16:02
I played around with this a bit and this very subtle limit may be involved: math.stackexchange.com/questions/160248/…. –  Jonathan May 18 '13 at 16:11
It's kind of odd integration limit. Why $[0,e]$ and not $[0,1]$? It should be clear that, for odd $n$: $$\int_0^e dx \log^n{x} = \int_0^{1/e} dx \log^n{x}$$ –  Ron Gordon May 18 '13 at 16:14
show 1 more comment

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.

-
You have a few mistakes. First, you can't just say the $e$ becomes unimportant without justification. While $a_n$ does increase towards infinite, at low values of $n$ the $e$ term matters a lot and this reflects on the sequence later. Also, $\left((-1)^n\right)^n = (-1)^{2n} = 1$. –  Jon Claus May 18 '13 at 19:13
You're right, that was a really dumb mistake on my part. Also, I did make a mistake in that I forgot $I_n = u_n e$ in my post, as opposed to $u_n$. So the limit is indeed $(-1)^{n^2}$. –  Jon Claus May 18 '13 at 19:17
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 $$\sum_{n = 1}^\infty \left(\frac{1}{n!}\int_0^e \log^n x \ dx\right)^n - (-1)^n$$