Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $5$?

share|improve this question
1  
Your sum is from $n=0$, so you should not factor out $a$.... –  N. S. Dec 18 '12 at 16:40
3  
"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
add comment

1 Answer

up vote 4 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 $n = 0 \implies r^0 = 1$.

share|improve this answer
    
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
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.