# Sums of Series

1. Jul 27, 2008

### Lance WIlliam

$$\sum$$ as n=0(theres a infinity above that sigma) , 3(1/11)^n

I thought it would just be 3/11 and converge due to the geometric test but its not....to find the sum would I just start at 0 and put numbers in for "n"...

2. Jul 27, 2008

### Defennder

You mean $$\sum^{\infty}_{n=0} 3\left(\frac{1}{11^n}\right)$$. You can factor the 3 outside of the sigma. And the resulting would be a geometric series, no?

3. Jul 27, 2008

### Lance WIlliam

yes but I dont see how I would go about finding a actual sum.

4. Jul 27, 2008

### Defennder

What do you know about the sum of a geometric series?

5. Jul 27, 2008

### Lance WIlliam

OH! S=a/1-r

6. Jul 27, 2008

### Lance WIlliam

its 33/10 thankyou