If $|a|<1$ then $\lim_{n \to \infty} a^n=0 $
I used to prove it by taking the logarithm but now saw a different solution in the book:
Consider the sequence $b_n = |a|^n$ , $b_{n+1}=b_n \cdot |a|$ < $b_n$ which means that $b_n$ is decreasing.
$b_1>b_2>....>b_n>0$ converges
Assume that $b_n$→$l$ , $b_{n+1} = |a| \cdot b_n$
Then, $b_{n+1} \to l$ and $b_n \to l \implies l=|a| \cdot l \implies l(1-|a|)=0 \implies l = 0$ as $1-|a|\neq0$.
I could not understand how he could write $b_{n+1} \to l$. Could you please explain it to me? Can we always make an assumption that if a series $c_n$ converges to $m$ then $c_{n+m}$, $m\in\mathbb N$ also converges to $m$ or $c_{n-1}$ also converges to $m$?