# What is a sample problem about finding the sum of a geometric sequence?

Apr 18, 2015

A geometric sequence is a sequence that involves multiplication to achieve a sequence. It is defined by:

${a}_{n} = {a}_{0} {r}^{n - 1}$ $\forall$ $n \ge 1$ and ${a}_{0} = a '$,
where $a '$ is some arbitrary first number or function and $r$ is some constant multiplier.

If you applied this to one like this:
$1 , 2 , 4 , 8 , \ldots$
you'd get:

${a}_{n} = 1 \cdot {2}^{n - 1}$ $\forall$ $n \ge 1$, ${a}_{0} = 1$, and $r = 2$.

A fuller example...
If you looked at this:
$1 , \frac{1}{2} , \frac{1}{4} , \frac{1}{8} , \frac{1}{16} , \ldots$
you can see that it's multiplying by $\frac{1}{2}$.

${a}_{n} = 1 \cdot {\left(\frac{1}{2}\right)}^{n - 1}$ $\forall$ $n \ge 1$, ${a}_{0} = 1$, and $r = 2$.

$\forall$ means "for all [in the set]".

Now if you wanted to write out a sum, you could begin with the power series:
${\sum}_{n = 0}^{N} {x}^{n} = 1 + x + {x}^{2} + {x}^{3} + \ldots$
which happens to equal:
$\frac{1}{1 - x}$

Then you can substitute in $\frac{1}{2}$ since $\frac{1}{2}$ was $r$:

$\implies 1 + \left(\frac{1}{2}\right) + {\left(\frac{1}{2}\right)}^{2} + {\left(\frac{1}{2}\right)}^{3} + \ldots$
$\implies 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots$
$\implies \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$

So it sums, and converges to 2. This, by the way, proves Zeno's Paradox false. You can take an infinite number of steps with halved distances and travel twice your initial distance.

$1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \ldots$
${\sum}_{n = 0}^{N} {\left(- 1\right)}^{n} {x}^{n} = {\sum}_{n = 0}^{N} {\left(- x\right)}^{n} = 1 - x + {x}^{2} - {x}^{3} + \ldots$
$\frac{1}{1 - \left(- x\right)} = \frac{1}{1 + x}$
And here, the sum is this if x = $\frac{1}{2}$:
$\frac{1}{1 - \left(- \frac{1}{2}\right)} = \frac{1}{\frac{3}{2}} = \frac{2}{3}$