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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 $5$?

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Your sum is from $n=0$, so you should not factor out $a$.... – N. S. Dec 18 '12 at 16:40
"using $a/(1-r)$" only works for series of the form $\sum_{n=0}^\infty a r^n$, when $0<|r|<1$. The given series is already in this form (with $a=1$). It might be better to memorize the sum of a convergent geometric series via "the first term of the series divided by $1-r$". – David Mitra Dec 18 '12 at 16:42
Right right! How could I forget. Thank you for clearing that up. Post it as an answer if you'd like so I can accept it. – Ceelos Dec 18 '12 at 16:45
up vote 3 down vote accepted

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 $n = 0 \implies r^0 = 1$.

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One takes $0^0=1$ for the formula above. – David Mitra Dec 18 '12 at 17:09
Why do you think that? I can't see it... – DonAntonio Dec 18 '12 at 18:17
@DonAntonio - are you asking me? – amWhy Dec 18 '12 at 18:19
Sorry, I was asking @DavidMitra – DonAntonio Dec 18 '12 at 22:30
@DonAntonio If the formula $(*)$ is to be true for $r=0$, then one has to take $0^0=1$ (just in interpreting this formula, of course). – David Mitra Dec 18 '12 at 23:05

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