1
$\begingroup$

If you have $\sum_{n = 0}^\infty(4/5)^n$ and you are asked to represent it as a geometric series you would:

$\sum_{n = 0}^\infty(4/5)(4/5)^{n-1}$ //factor out your constant
therefore $a = 4/5$, $r = 4/5$, $|r| < 1$ checks out.
Using $a / (1 - r)$ you get $(4/5)/(1 - 4/5)$
$(4/5)/(1/5) = 4$. which confuses me because the solution said the answer was $5$?

  • 0
    Right right! How could I forget. Tha$n$k you for cleari$n$g that up. Post it as an answer i$f$ you'd like so I can accept it.2012-12-18

1 Answers 1

3

You have $\sum_{n = 0}^\infty \left(\frac45\right)^n$

We use the fact that:

$\text{If}\;\;0 < r < 1,\,\text{ then}\;\;\sum_{n=0}^\infty r^n = \dfrac{1}{1- r}\tag{*}$

So we have $r = \dfrac{4}{5} < 1$.

$\sum_{n = 0}^\infty \left(\frac{4}{5}\right)^n = \frac{1}{1 - (4/5)} = 5.$ Note: the "$1$" in the numerator of (*) can be thought of as the first term of the sum: $r^n$ at $n = 0 \implies r^0 = 1$.

  • 0
    Then again, $0^0=1$ is simply true (cardinality of the set of functions from the empty set to itself). It's merely the problem that $x\to0$ and $y\to 0$ does not imply $x^y\to 1$.2012-12-19