# limit of geometric sequence

As mentionned in the geometric sequence entry,

 $\displaystyle\lim_{n\to\infty}ar^{n}=0$ (1)

for  $|r|<1$.  We will prove this for real or complex values of $r$.

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

 $s^{n}=(1+s-1)^{n}>1^{n}+\binom{n}{1}(s-1)=1+n(s-1)>n(s-1)>M$

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

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

 $|r^{n}|=|r|^{n}=\frac{1}{s^{n}}<\frac{1}{n(s-1)}<\varepsilon$

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

 $\lim_{n\to\infty}r^{n}=0,$

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

Title limit of geometric sequence LimitOfGeometricSequence 2013-03-22 18:32:43 2013-03-22 18:32:43 pahio (2872) pahio (2872) 6 pahio (2872) Proof msc 40-00