# Thread: sum of infinite series

1. ## sum of infinite series

this is for my stats class and i dont have a calc book with me...

the sum of 1/(3^x) where x is from 1 to infinity

thanks

2. $\frac{a}
{{1 - r}} = \frac{{\frac{1}
{3}}}
{{1 - \frac{1}
{3}}}$

3. what do a and r represent, and why are they both 1/3 in this case? thanks

4. a is the first term of the series, and r is the common ratio (the ratio between consecutive terms).

5. This is a geometric series so you use the formula plato gave to find the sum

6. ok what if the ratio is 5x/6, is there another formula i can use?

this is for solving the sum of:

x * (1/6) * (5/6)^(x-1)

where x is 1 to infinity

7. $\sum\limits_{k = J}^\infty {ax^k } = \frac{{ax^J }}
{{1 - x}},\,\,\left| x \right| < 1$

8. Originally Posted by Plato
$\sum\limits_{k = J}^\infty {ax^k } = \frac{{ax^J }}
{{1 - x}},\,\,\left| x \right| < 1$

does a = x*(1/6) then?

that wont work because then the answer still has an x in it.

9. Originally Posted by Dubulus
ok what if the ratio is 5x/6, is there another formula i can use? this is for solving the sum of:
x * (1/6) * (5/6)^(x-1) where x is 1 to infinity
To be honest, I cannot say that I really follow what you are trying to do.
Here is an example: $\sum\limits_{x = 1}^\infty {\left( {\frac{5}{6}} \right)^x } = \frac{{\frac{5}{6}}}{{1 - \frac{5}{6}}}$

10. Originally Posted by Dubulus
ok what if the ratio is 5x/6, is there another formula i can use?

this is for solving the sum of:

x * (1/6) * (5/6)^(x-1)

where x is 1 to infinity
The correct formula for this particular sum is $\frac{a}{(1-r)^2}$.

11. Originally Posted by icemanfan
The correct formula for this particular sum is $\frac{a}{(1-r)^2}$.

my problem is that there is an "x" in the sum. this to me means that each new factor in the series isn't changing by a constant ratio.