# Limit proof of $a^n$ where |a|<1

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 $m$?

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$\sum a^n = 1 / (1-a)$. On the other hand $\lim a^n =0$. – Nikita Evseev Dec 5 '12 at 3:36
@nikita2 you are right. I edited my question. – Amadeus Bachmann Dec 5 '12 at 3:38

Well, since we're letting the parameter $n$ get arbitrarily large, there's no difference in saying $n$, $n+1$, $n+m$ for some $m > 0$, or even saying $2n$, $3n$, etc.

The reasoning behind this is that the only thing we care about when dealing with the limit of a sequence is how it behaves asymptotically i.e. when $n$ is very large.

A bit more rigor. A sequence $\{a_n\}_{n \in \mathbb{N}}$ approaches a finite limit $l$ if for any $\varepsilon > 0$ there exists $N = N(\varepsilon) \in \mathbb{N}$ such that for any $n > N$, $|a_n-l|<\varepsilon$. Now suppose $a_n \to l$, and let $b_n = a_{n + m}$. Let $\varepsilon > 0$ be given, and we know that $|a_n - l| < \varepsilon$ for any $n > N = N(\varepsilon)$. Now, we know that $n+m > n > N$, so $|b_n - l | = |a_{n+m} - l|<\varepsilon$, so $b_n \to l$ as well.

To answer your second question, will this work if $m$ is negative? Yes, if $a_n \to l$ and $b_n = a_{n+m}$, then $b_n \to l$ as well. Take $N_b$ to be equal to $N_a + m$, and the condition is satisfied.

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If I understand your question, to say $c_{n} \rightarrow a$ is to say that for every $\epsilon>0$, there exists an $N$ so that for every $n>N$, $|c_{n}-a|<\epsilon$. To say that $c_{n+1} \rightarrow a$ is to say that for every $\epsilon>0$ there exists an $M$ so that for every $n+1>M$, $|c_{n+1}-a|<\epsilon$. Can you see how the two statements are equivalent?

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yeah it is so simple. how could not I think that! :/ I guess I need to take a rest. thank you for your answer. – Amadeus Bachmann Dec 5 '12 at 3:55