Convergence of Geometric Series

Key Questions

How can I tell whether a geometric series converges?
Answer:

A geometric series of geometric sequence $u_{n} = u_{1} \cdot r^{n - 1}$ $u_n= u_1 * r^(n-1)$ converges only if the absolute value of the common factor $r$ $r$ of the sequence is strictly inferior to $1$ $1$ ; in other words, if $| r | < 1$ $|r|<1$ .
Explanation:
The standard form of a geometric sequence is :

$u_{n} = u_{1} \cdot r^{n - 1}$ $u_n = u_1 * r^(n-1)$

And a geometric series can be written in several forms :

$+ \infty \sum n = 1 u_{n} = + \infty \sum n = 1 u_{1} \cdot r^{n - 1} = u_{1} + \infty \sum n = 1 r^{n - 1}$ $sum_(n=1)^(+oo)u_n = sum_(n=1)^(+oo)u_1*r^(n-1) = u_1sum_(n=1)^(+oo)r^(n-1)$

$= u_1*lim_(n->+oo)(r^(1-1) + r^(2-1) + r^(3-1) + ... + r^(n-1))$

Let $r_n = r^(1-1) + r^(2-1) + r^(3-1) + ... + r^(n-1)$

Let's calculate $r_n - r*r_n$ :

$r_n - r*r_n = r^(1-1) - r^(2-1) + r^(2-1) - r^(3-1) + r^(3-1) + ... - r^(n-1) + r^(n-1) - r^n = r^(1-1) - r^n$

$r_n(1-r) = r^(1-1) - r^n = 1 - r^n$

$r_n = (1 - r^n)/(1-r)$

Therefore, the geometric series can be written as :

$u_1sum_(n=1)^(+oo)r^(n-1) = u_1*lim_(n->+oo)((1 - r^n)/(1-r))$

Thus, the geometric series converges only if the series $sum_(n=1)^(+oo)r^(n-1)$ converges; in other words, if $lim_(n->+oo)((1 - r^n)/(1-r))$ exists.

If |r| > 1 : $lim_(n->+oo)((1 - r^n)/(1-r)) = oo$

If |r| < 1 : $lim_(n->+oo)((1 - r^n)/(1-r)) = 1/(1-r)$ .

Therefore, the geometric series of geometric sequence $u_n$ converges only if the absolute value of the common factor $r$ of the sequence is strictly inferior to $1$ .
Shura · 2 · Jul 2 2015
What are some examples of infinite geometric series?

Answer:

Here are some examples:

$1 + 1/2 + 1/4 + 1/8 + 1/16 +...$

$1 - 1 + 1 - 1 + 1 - 1 +...$

$1 + 2 + 4 + 8 + 16 +...$

Explanation:

All geometric series are of the form $sum_(i=0)^oo ar^i$ where $a$ is the initial term of the series and $r$ the ratio between consecutive terms.

In the three examples above, we have:

$a = 1$ , $r = 1/2$

$sum_(i=0)^oo ar^i = 2$

$a = 1$ , $r = -1$

$sum_(i=0)^oo ar^i$ does not converge - it alternates between $0$ and $1$ as each term is added.

$a = 1$ , $r = 2$

$sum_(i=0)^n ar^i -> oo$ as $n->oo$

The geometric series $sum_(i=0)^oo ar^i$ only converges in the following cases:

(1) $a = 0$

$sum_(i=0)^oo ar^i = 0$

(2) $abs(r) < 1$

$sum_(i=0)^oo ar^i = a/(1-r)$

George C. · 1 · Jul 13 2015
What is the geometric power series?

$a+ax+ax^{2}+ax^{3}+cdots$ , which converges to $a/(1-x)$ when $|x|<1$ . More generally, you could also write $a+a(x-c)+a(x-c)^{2}+a(x-c)^{3}+cdots$ , which converges to $a/(1-(x-c))=a/((1+c)-x)$ when $|x-c| < 1$ .

Bill K. · 1 · Dec 7 2015

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