Summing up the Infinite: 3(1/11)^n

[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"...

 

[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?

 

[Lance WIlliam]

yes but I don't see how I would go about finding a actual sum.

 

[Defennder]

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

 

[Lance WIlliam]

OH! S=a/1-r

 

[Lance WIlliam]

its 33/10 thankyou