Prove $\sum\limits_{n=1}^{\infty} \frac{1}{\int_0^1\ln^n(x)dx} = \frac{1}{e}-1$ Ok, that's really not an easy question to convey in text. What I'm asking is this:
$$\sum\limits_{n=1}^{\infty} \frac{1}{\int_0^1\ln^n (x)dx} = \frac{1}{e}-1$$
Numerically, I'm pretty sure it does (it's about -0.63212). But, I have no idea how I'd go about proving it.
Any insights?

Note
$$\int_{0}^{1}\ln^nx dx = \frac{d^n}{da^n}\left(\int_{0}^{1} x^a dx\right)\bigg|_{a=0}
= \frac{d^n}{da^n}\left(\frac1{a+1}\right) \bigg|_{a=0} =(-1)^{n}n!
$$
Thus
$$\sum\limits_{n=1}^{\infty} \frac{1}{\int\limits_0^1 \ln^nx  dx} = 
\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n!}
=\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{n!}-1=
e^{-1}-1$$
