Let $\theta$ denote a smoothly distributed random variable with support $[0, 1]$. I am trying to evaluate
$$ \lim_{n \rightarrow \infty} \frac{\mathbb{E}[\theta^n]}{\mathbb{E}[\theta^{n-1}]}$$
I suspect, but cannot show, that the limit equals $1$. Does anyone know how to do this?
My attempts so far: Since $\theta \in [0, 1]$, it seems reasonably clear that both $\mathbb{E}[\theta^n] \rightarrow 0$ and $\mathbb{E}[\theta^{n-1}] \rightarrow 0$ as $n \rightarrow \infty$ (we are raising numbers that are less than $1$ to ever higher powers). Thus, we can apply L'Hopital's rule to find that
$$ \lim_{n \rightarrow \infty} \frac{\mathbb{E}[\theta^n]}{\mathbb{E}[\theta^{n-1}]} \equiv \lim_{n \rightarrow \infty} \frac{\int_0^1 \theta^nf(\theta)d\theta}{\int_0^1 \theta^{n-1}f(\theta)d\theta} = \lim_{n \rightarrow \infty} \frac{\int_0^1 \ln(\theta)\theta^nf(\theta)d\theta}{\int_0^1 \ln(\theta)\theta^{n-1}f(\theta)d\theta}$$
I am a bit unclear, however, how to proceed from this point (or whether better approaches are available).
We have
$$ \mathbb{E}[\theta^n]^{\frac{n+1}{n}} \leq \mathbb{E}[\theta^{n+1}] \leq \mathbb{E}[\theta^n]. $$
Indeed, the first inequality is the consequence of the Jensen's inequality and the second inequality follows from $\mathbb{P}(\theta\in[0,1])=1$. Dividing each side by $\mathbb{E}[\theta^n]$, we get
$$ \mathbb{E}[\theta^n]^{1/n} \leq \frac{\mathbb{E}[\theta^{n+1}]}{\mathbb{E}[\theta^n]} \leq 1. $$
Now by noting that $\mathbb{E}[\theta^n]^{1/n} \to \| \theta \|_{\infty} = 1$ as $n\to\infty$ by the assumption, the desired conclusion follows.