limit of geometric sequence
As mentionned in the geometric sequence entry,
limn→∞arn=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
limn→∞sn=∞ (cf. limit of real number sequence). In fact, if M is an arbitrary positive number, the binomial theorem (or Bernoulli’s inequality) implies that
sn=(1+s-1)n>1n+(n1)(s-1)=1+n(s-1)>n(s-1)>M |
as soon as n>Ms-1.
Let now |r|<1 and ε be an arbitrarily small positive number. Then |r|=1s with s>1. By the above remark,
|rn|=|r|n=1sn<1n(s-1)<ε |
when n>1(s-1)ε. Hence,
limn→∞rn=0, |
which easily implies (1) for any real number a.
Title | limit of geometric sequence |
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Canonical name | LimitOfGeometricSequence |
Date of creation | 2013-03-22 18:32:43 |
Last modified on | 2013-03-22 18:32:43 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 6 |
Author | pahio (2872) |
Entry type | Proof |
Classification | msc 40-00 |