limit of geometric sequence

As mentionned in the geometric sequence entry,

$\underset{n\to \mathrm{\infty }}{lim}a{r}^n=0$(1)

for  .  We will prove this for real or complex values of $r$.

We first remark, that for the values  $s>1$  we have $\underset{n\to \mathrm{\infty }}{lim}{s}^n=\mathrm{\infty }$ (cf. limit of real number sequence).  In fact, if $M$ is an arbitrary positive number, the binomial theorem (or Bernoulli’s inequality) implies that

$$s^n={(1+s-1)}^n>1^n+\left(\genfrac{}{}{0pt}{}{n}{1}\right)(s-1)=1+n(s-1)>n(s-1)>M$$

as soon as  $n>{\displaystyle \frac{M}{s-1}}$.

Let now    and $\epsilon$ be an arbitrarily small positive number.  Then  $|r|={\displaystyle \frac{1}{s}}$  with $s>1$.  By the above remark,

when  $n>{\displaystyle \frac{1}{(s-1)\epsilon }}$.  Hence,

$$\underset{n\to \mathrm{\infty }}{lim}{r}^n=0,$$

which easily implies (1) for any real number $a$.