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Show that the sequence {$r^n$} converges to $0$ if $|r|<1$.

My attempt

Let $a_{n} = r^n$ and $\varepsilon>0$.

$|a_{n} - 0| = |r^n| = |r|^n$.

Consider $|r|^n < \varepsilon$.

Taking log on both sides we get,

$n$log$|r|<$log$\varepsilon$.

Since $|r| = \frac{1}{1+h}, h>0$, log$|r|$ is negative. Therefore,

$n>\frac{log\varepsilon}{log(\frac{1}{1+h})}$.

Let $m$ be any positive integer greater than $\frac{log\varepsilon}{log(\frac{1}{1+h})}$. Then for $\varepsilon>0$, there exists a positive integer $m$ such that $|a_{n} - 0|<\varepsilon$ for all $n\geq$ $m$.

Therefore, the sequence {$r^n$} converges to $0$ if $|r|<1$.

Is this method correct? Most proofs I saw make use of binomial expansion and all, and I wonder why.

Mittens
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Sasikuttan
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    To answer your question in the end: in the calculus course which I teach, this result is proved before logarithm (and the number $e$ as well) is formally defined. So using your proof at the beginning of the course would be kind of cheating. This is why we use more basic tools, like the binomial theorem. – Mark Jan 24 '22 at 16:26
  • @Mark I see. Good to know that. – Sasikuttan Jan 24 '22 at 16:28

3 Answers3

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Beginning the proof with “Take $|r|^n<\varepsilon$” makes no sense, since the goal is to prove that $\lim_{n\to\infty}|r|^n=0$. You should begin with something like “Let $\varepsilon>0$; I will prove that there is some $M\in\Bbb N$ such that $|r|^n<\varepsilon$ if $n\geqslant m$.”

The rest is correct, but it can be made simpler. Since $|r|<1$ and since $\log|r|$ is such that $e^{\log|r|}=|r|<1$, $\log|r|<0$. No need to write $r$ as $\frac1{1+h}$.

José Carlos Santos
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Be careful with the wording of your proof. What you want to do is show that for a given $\epsilon$, the inequation $|r|^n\lt \epsilon$ is equivalent to $n\gt -\log(\epsilon)/\log(1+h)$. As you have written it here, you only show the direct implication, which is of no use to you.

With these corrections, the proof you give is correct, as far as I can tell.

Astyx
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By a known theorem, if $$\lim_{k \rightarrow \infty } \sum^{k}_{i=0} a_k $$ exists, $\{a_k\}$ converges to zero. Now, by the geometric series sum, we have that the sequence $\{s_k\}$ given by $$ s_k := \sum^{k}_{i=0} r^{i},$$ for all $k\in \mathbb{N},$ is an increasing and bounded sequence (bounded below by zero and above by $\dfrac{1}{1-r}$). This means that $\{s_k\}$ converges, by the Monotone convergence theorem, and, hence, $r^{k}\rightarrow 0$ as $k\rightarrow \infty.$

R. W. Prado
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  • I like your thinking! That said, this argument feels circular because the sum of the geometric series is usually established with the identity $\sum_{n=0}^k r^n=\frac{1-r^{k+1}}{1-r}$ and the fact that $\lim_{n\to\infty}r^n=0$ whenever $|r|<1$, which is what we're trying to prove. – Alann Rosas Jan 25 '22 at 00:32
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    It feels circular, but it is not. Since, $$\sum^{k}{i=0} r^{k} = \dfrac{1-r^{k+1}}{1-r},$$ this means that $$\sum^{k}{i=0} r^{k} \leq \dfrac{1}{1-r}.$$ Thus, ${s_k}$ is bounded above (it's automatically bouded below by zero as sum of positive terms). It's not needed to know the limit of a geometric sum to prove the geometric series is bounded above. Its convergence can be established without knowing it's limit. After that, as the series converges, its terms $r^{k}$ converges to zero. – R. W. Prado Jan 25 '22 at 20:47