Is this alternate proof of convergence of geometric series correct? 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.
 A: 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 $\frac1{1+h}$.
A: 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.
A: 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 $k\rightarrow \infty.$
