Limit of the geometric sequence $\{r^n\}$, with $|r| < 1$, is $0$?

Question

Prove that the $\lim_{n\to \infty} r^n = 0$ for $|r|\lt 1$.

I can't think of a sequence to compare this to that'll work. L'Hopital's rule doesn't apply. I know there's some simple way of doing this, but it just isn't coming to me. :(

Answer 1

Let $a_n=|r|^n$ for $n\ge1$, so $(a_n)$ is a decreasing sequence which is bounded below by zero

and therefore converges, so let $\displaystyle\lim_{n\to\infty}a_n=L$.

Since $a_{n+1}=|r|a_n$, and $(a_{n+1})$ is a subsequence of $(a_n)$, by the theorem that all subsequence of a convergent sequence converge to the same limit$, \;\;L=|r|L\implies L=0\;\;$ (for $r\ne0$).

Since $\displaystyle\lim_{n\to\infty}|r|^n=0$, $\displaystyle\lim_{n\to\infty}r^n=0$ also.

Answer 2

The case where $r=0$ is trivial. WLOG, suppose that $0<r<1$.

We let $M= \frac{1}{r}-1$.

Now take any $\epsilon >0$. There exists $N \in \mathbb{N}$ such that $N > \frac{1}{\epsilon M}$.

By Binomial expansion, for $n \geq N$,

\begin{eqnarray} r^n &= & \frac{1}{(1+M)^n} \\ & \leq & \frac{1}{1+nM} \\ & \leq & \frac{1}{nM} \\ & \leq & \frac{1}{NM} \\ & < & \epsilon. \end{eqnarray}

Answer 3

I'll assume $0 < |r| < 1$.

We want $|r^n| < \epsilon$, or equivalently $(1/|r|)^n > 1/\epsilon$.

Write $1/|r| = 1 + a$ with $a > 0$. Then $(1/|r|)^n = (1+a)^n \geq 1 + na$ by the binomial theorem.

Choose $N$ such that $1 + Na > 1/\epsilon$, say let $N = \lceil \frac{1}{a\epsilon}\rceil$. Then $|r^n| < \epsilon$ for $n \geq N$.

Answer 4

Let $u=1-|r|$ and note that $|r|\lt1$ implies $0\lt u\le1$. This implies $0\le1-u^2\lt1$, which in turn implies $0\le1-u\lt1/(1+u)$, which in turn implies the first (strict) inequality in the string of assertions

$$|r^n-0|=|r|^n=(1-u)^n\lt{1\over(1+u)^n}\le{1\over1+nu}\lt{1\over nu}=\epsilon{1\over nu\epsilon}\le\epsilon{\lceil1/(u\epsilon)\rceil\over n}$$

for all $n\ge1$ and any $\epsilon\gt0$. So letting $N_\epsilon=\lceil1/(u\epsilon)\rceil$, we find that $n\ge N_\epsilon$ implies $|r^n-0|\lt\epsilon$, which tells us $\lim_{n\to\infty}r^n=0$.

Remark: This answer follows much the same logic as those of aes and Richard, differing mainly in style of presentation. Its one advantage, if any, is that it does not split out the trivial case of $r=0$. The key inequality, $(1+u)^n\ge1+nu$, can, if need be, be proved by induction. The other equalities and inequalities in the display are straightforward to verify; in particular, the equal sign preceding the sudden appearance of $\epsilon$ is justified by the fact that all we're doing there is multiplying and dividing by a nonzero quantity.

Answer 5

If $r=0$ it's trivial, so we can skip this case. Now assume $r\neq0$.

$$\lim_{n\to\infty}r^n=0 \Longleftrightarrow(\forall \epsilon \in \mathbb{R}^+)(\exists \delta\in\mathbb{R})(\forall n_{> \delta})(|r^n| < \epsilon)$$

Now we can note, that both sides of above inequality are positive. We can use logarithms.

$$|r^n|<\epsilon \overset{|r|<1}{\Longleftrightarrow} n > \log_{|r|}\epsilon $$

Because $\log_{|r|}\epsilon$ is const, $\delta = \log_{|r|}\epsilon$ satisfy thesis.

$\mathscr{Q.E.D.}$

Answer 6

For fun, I'll try to do a proof by contradiction.

Suppose $\lim_{n \to \infty} r^n \ne 0 $. Then, since $S =(r^n)_{n=1}^{\infty}$ is a decreasing sequence (since $0 < r < 1$), $s = \lim S$ exists and $s > 0$. (Only $\lim\inf$ is needed for what follows.)

Then, for any $c > 0$, there is an $n(c)$ such that $r^{n(c)} < s+c $.

Now, consider $r^{n(c)+1}$. $r^{n(c)+1} =r^{n(c)}r <(s+c)r $. But this can be made less than $s$ if $(s+c)r < s $ or $sr+cr < s$ or $cr < s-sr = s(1-r)$ or $c < \frac{s(1-r)}{r}$. But, by assumption, $r^{n(c)+1} \ge s $.

Therefore the assumption that $\lim r^n > 0$ leads to a contradiction, so $\lim r^n = 0$.

Answer 7

This is an answer I had earlier posted.

This is not a rigorous proof but helps in visualisation.

If $$-1<x<1$$ and you raise it to any power $n > 1 $ the value of it decreases.

Eg say$ x = 0.1 , x ^2 = 0.01, x^3 = 0.001$ or $x ^n = 0.1 ^ n = \frac{1}{10^n}$ and as $ n$ keeps getting bigger , the denominator approaches $\infty $ and the numerator approaches $0 $