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

Solution 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.

Solution 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}

Solution 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 $n \geq N$.

Solution 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.