1. ## series (serious!)

determine the values of the series <E(thats sposed to be a sigma) k=0 to oo (1/2)^k

so what do i actually need to do because it goes up to infinity, you cant just solve each term and add them, what is the startergy to doing this? thank you kindle

2. $\displaystyle \sum_{k = 0}^{\infty}\left(\frac{1}{2}\right)^k$ is a geometric series with $\displaystyle a = 1$ and $\displaystyle r = \frac{1}{2}$.

As long as $\displaystyle |r| < 1$ (as it is in this case) the infinite sum of a geometric series is equal to $\displaystyle \frac{a}{1- r}$.

3. Originally Posted by mathcore

determine the values of the series <E(thats sposed to be a sigma) k=0 to oo (1/2)^k

so what do i actually need to do because it goes up to infinity, you cant just solve each term and add them, what is the startergy to doing this? thank you kindle
It is a convergent geometric series, you use the formula for the sum in your notes and/or text.

CB

4. $\displaystyle \sum_{k=0}^{\infty}\left(\frac{1}{2}\right)^k = \sum_{k=0}^{\infty}x^k\bigg|_{x=\frac{1}{2}} = \frac{1}{1-x}\bigg|_{x=\frac{1}{2}} = \cdots$

5. i dont have "notes and/or text" as u put it

does it make a difference whether k = 0 to infinite or k=1 to infite? why is a =1?

also, is a=1 because thats the first term (as its a zero power). if its a different first term, do you just dddddddddd

6. Yes, $\displaystyle a = 1$ and the sum $\displaystyle = 2$.

7. Originally Posted by mathcore
yes, the sum of the infinite geometric series $\displaystyle \sum_{k = 0}^{\infty} \left(\frac{1}{2}\right)^k = 2$
$a = 1$ because the first term of the series is $\displaystyle \left(\frac{1}{2}\right)^0 = 1$.
if the series started with $k = 1$ , then $\displaystyle a = \frac{1}{2}$