A sequence $(a_n)$ converges to $L$, $|L| < 1$. Prove that $(a_n)^n$ converges to $0$ So $a_n$ converges to $L$, and the absolute value of $L$ is less than $1$. How do we go about proving that the sequence $(a_n)^n$ converges to 0? I tried a couple of basic methods but I don't seem to have made any progress.
 A: Hint: $|a_n^n| \le \left(\dfrac{|L|+1}{2}\right)^n$ for sufficiently large $n$.

 $a_n \to L \quad \Longrightarrow \quad \forall \varepsilon>0, \exists N_{\varepsilon} \in \mathbb{N}, \forall n>N_{\varepsilon} \colon |a_n-L| < \varepsilon \quad \Longrightarrow \quad |a_n| < |L|+\varepsilon,\ \forall n>N_{\varepsilon}$.
 Now let $\varepsilon := \dfrac{1-|L|}{2}>0$.

A: If $a_n$ converges to $L$ then $|a_n|$ will converge to $|L|$ and therefore $\log|a_n|$ will converge to $\log|L|$. For $\epsilon\leq \frac{-\log|L|}{2}$, we can find $N_0$, such that for $n>N_0$ :
$$
\log|a_n|\leq ({\log|L|+\epsilon})=\frac{\log|L|}{2}
$$
So we have:
$$
\log|a_n|^n\leq n\left(\frac{\log|L|}{2}\right)
$$
Because $\alpha=\frac{\log|L|}{2}<0$ we have:
$$
|a_n|^n\leq e^{\alpha n}
$$
And the right side goes to zero as $n\rightarrow \infty$.
A: Given $\varepsilon \in (|L|,1)$ there exists some $N \in \mathbb{N}$ such that
$$
|a_n-L|\le \varepsilon-L \quad \forall n \ge N.
$$ 
Therefore for every $n \ge N$ we have
$$
|a_n^n|=|a_n|^n \le (|a_n-L| +L)^n\le \varepsilon^n.
$$
Since $0<\varepsilon<1$, it follows that
$$
\lim_n|a_n^n| \le 0, 
$$
i.e. $\lim_na_n^n=0$.
