# Summation ?

1. Aug 20, 2008

### Wholewheat458

1. The problem statement, all variables and given/known data
I am to find the sum of the series, but what do i do if it is infinite?? no clue.
i'm also not sure how to type the symbols so i hope you can understand me:shy:
: (Sum) n=0, limit = infinity: 1/2(2/3)^n

2. Relevant equations
i 'm not sure.

3. The attempt at a solution
i understand how to add up to a limit, but what do i do with the infinity?

2. Aug 20, 2008

### snipez90

Well if the 1/2 is the first term in the product then you can pull it out front (this is obvious by the distributive law). Then recognize it as a geometric series. There is a simple formula for an infinite geometric series which you should try deriving yourself.

3. Aug 20, 2008

### Wholewheat458

.. 1/2(sum)n=0, infinity (2/3)^n+1 ..
?? do you just leave it like that?..
as a sort of equation?

4. Aug 20, 2008

### snipez90

Ok an infinite geometric series like a normal one has a first term conventionally detonated as $$a$$ and common ratio $$r$$. Since we start at n = 0, the first term is (2/3)^0 = 1 (You can't just change the exponent from n to n+1 unless you change the starting value. Making do with what you're given is best in this case). Then we go to n = 1, the second term is (2/3)^1 = 1*(2/3) = 2/3. Then for n = 2, the third term is (2/3)^2 = (2/3)(2/3). Now the idea is to sum all of these terms, i.e. 1 + 1*(2/3) + 1*(2/3)*(2/3) + 1*(2/3)(2/3)(2/3) + ...

So we can generalize a bit. We have $$a$$ as our first term and a common ratio $$r$$. Our sum, which we'll denote $$S$$ is

$$S = a + ar + ar^2 + ar^3 + ...$$ (1)

(Compare with the first paragraph to understand why this is true).

Now we need to solve for $$S$$ because that gives us the sum. The trick here is to multiply $$S$$ by our common ratio $$r$$, i.e. multiply both sides of (1) above by $$r$$ and write it underneath. Then subtract the new equation from (1) and see for yourself how all the terms cancel and allow you to solve easily for $$S$$

5. Aug 20, 2008

### tiny-tim

Hi Wholewheat458!

$$\sum_{n=0}^{\infty} \frac{1}{2}\,\left(\frac{2}{3}\right)^n$$

$$=\ \frac{1}{2}\,\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n$$

You should know what $$\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n$$ is,

but if you don't, just sum it from 0 to N, and then let N --> ∞

(btw, it's [noparse]$$\sum_{n=0}^{\infty} \frac{1}{2}\,\left(\frac{2}{3}\right)^n$$[/noparse] )