A set of lecture notes that I'm reading has a proof which seems to use a result which could perhaps be stated as the following:
If $\{a_n\}$ is a sequence with $a_n\to L$ and $\vert L\vert<1$, then if the sequence $\{b_n\}$ is defined as $b_n=(a_n)^n$ for all $n$, then $b_n\to0$.
I had a look around online but couldn't find a similar statement, I don't know if this statement is true though I would expect that it is. My questions are: Is the statement true? Is my attempt at a proof correct? Here's what I've written:
Because $\vert L\vert<1$, the absolute values of the terms of $a_n$ must eventually be bounded strictly below $1$. More concretely choose $\varepsilon>0$ small enough so that $\vert L\vert +\varepsilon <1$ (possible since $\vert L\vert <1$), then because $a_n\to L$ there exists $N\in\mathbb{N}$ such that $$ \vert a_n-L\vert<\varepsilon\hspace{1cm}\forall n\geq N $$ and since $\vert a_n\vert-\vert L\vert\leq\vert a_n-L\vert$ we have $$ \vert a_n\vert < \vert L\vert + \varepsilon\hspace{1cm}\forall n\geq N $$ Set $c = \vert L\vert + \varepsilon$, then $$ -c^n<(a_n)^n<c^n\hspace{1cm}\forall n\geq N $$ So since $\vert c\vert<1$ we have $-c^n,c^n\to 0$ and so by the pinching (squeeze) theorem $b_n=(a_n)^n\to0$.