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[parent] limit of geometric sequence (Proof)

As mentionned in the geometric sequence entry,

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

for $ \vert r\vert < 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

$\displaystyle s^n = (1+s-1)^n > 1^n+\binom{n}{1}(s-1) = 1+n(s-1) > n(s-1) > M$
as soon as $ \displaystyle n > \frac{M}{s-1}$.

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

$\displaystyle \vert r^n\vert = \vert r\vert^n = \frac{1}{s^n} < \frac{1}{n(s-1)} < \varepsilon$
when $ \displaystyle n > \frac{1}{(s-1)\varepsilon}$. Hence,
$\displaystyle \lim_{n\to\infty}r^n =0,$
which easily implies (1) for any real number $ a$.



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Cross-references: implies, Bernoulli's inequality, binomial theorem, number, positive, limit of real number sequence, complex, real, geometric sequence

This is version 3 of limit of geometric sequence, born on 2008-11-19, modified 2008-11-19.
Object id is 11264, canonical name is LimitOfGeometricSequence.
Accessed 180 times total.

Classification:
AMS MSC: 40-00 (Sequences, series, summability :: General reference works )
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