Please help! Find the following infinite sum, if its exists. If one does not exist enter "no infinite sum."
(Look at response.)

Question

anonymous · Answer

$$\sum_{n-1}^{\infty}4(1/3)^(n-1)$$

anonymous · Answer

n-1 is raised (^)

anonymous · Answer

$$4\sum _{n=1}^{\infty } \left(\frac{1}{3}\right)^{n-1}$$

anonymous · Answer

I am going to change it

anonymous · Answer

$$4\sum _{n=0}^{\infty } \left(\frac{1}{3}\right)^n$$

anonymous · Answer

That's correct :/ Sorry

anonymous · Answer

This is formula
$$\sum _{n=0}^{\infty } a^n =\text{  }\frac{1}{1-r}$$

$$\sum _{n=0}^{\infty } (1/3)^n =\text{  }\frac{1}{1-(1/3)}$$

anonymous · Answer

1/(2/3)
3/2

anonymous · Answer

So the infinite sum would be 3/2?

anonymous · Answer

yes, by the way  we were able to do it because 1/3 is less than 1

zarkon · Answer

times 4

anonymous · Answer

Huh Zarkon?

zarkon · Answer

3/2*4=6

anonymous · Answer

oh yeah , we forgot about 4 infront

anonymous · Answer

Oh I see. So final answer= 6. Thank you guys! :)