# 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 $a$.

Title | limit of geometric sequence |
---|---|

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 |