Sigma notation problem

1. Sep 9, 2006

sammieh

Help!

We need to solve the following problem:

∞ r-1
find x if ∑ (3x/2) = 4
r=1

We know the answer is 1/2, but we have no idea how they got that answer. Can anyone help?

Thanks,
Sammie

2. Sep 9, 2006

sammieh

it looks like things shifted when we wrote our problem. It should be the summation of (3x/2) raised to the r-1 power = 4 where r is going from 1 to infinity.

I hope this makes sense to someone!

3. Sep 9, 2006

0rthodontist

Do you know the formula for an infinite geometric sum?

4. Sep 9, 2006

sammieh

Is it something like

Sn = u1(r^n – 1)
---------- where Sn is really S sub n and u1 is really u sub 1?
r-1

5. Sep 9, 2006

0rthodontist

What does that formula mean? Identify u1, r, and especially n.

6. Sep 9, 2006

sammieh

o.k

r is the constant common ratio and u1 is the first term of the series and I think n is a term in the series??

7. Sep 9, 2006

0rthodontist

Well, first step is to find out what n is. You have to know what your formulas mean before you'll be able to use them.

You might also have another formula that is more directly useful, but you should still understand that one.

Last edited: Sep 9, 2006
8. Sep 9, 2006

sammieh

o.k. n is the number of terms we are adding up. In our case infinity???

Right?

9. Sep 9, 2006

0rthodontist

Right--though it's not "really" infinity, you take the limit as n goes to infinity. What you have is really the formula for a finite sum, which you could tell from the n. So--if you don't have it already, what do you think the formula for an infinite sum is? In an infinite sum, r will be less than 1, so what is the limit of r^n as n goes to infinity?

Then it's just a matter of applying the formula to the series you gave and solving for x.

10. Sep 9, 2006

sammieh

i'm still confused. Is the formula for an infinite sum:

u1/r-1 ?

and is the limit 1 as r goes to infinity?

11. Sep 9, 2006

0rthodontist

n goes to infinity, not r--typo on my part. You're close, but not quite. As n goes to infinity, what does r^n go to? Then what does r^n - 1 go to?

12. Sep 9, 2006

sammieh

wouldn't r^n go to 1, but never reach 1 since it is always <1?

and if r^n goes to 1 then wouldn't r^n-1 go to 0?

13. Sep 9, 2006

0rthodontist

Pick some r < 1 and find out what it does. What is r^10? What is r^20?

14. Sep 9, 2006

sammieh

o.k for example if we take r=.5 then as you increase the exponent r gets closer & closer to 1. right?

15. Sep 9, 2006

0rthodontist

Well, does it?

16. Sep 9, 2006

sammieh

yes it does. as we increased the exponent the answer kept getting closer and closer to 1.

17. Sep 9, 2006

0rthodontist

What's 0.5^2?

18. Sep 9, 2006

sammieh

.5 ^ 2 = .25

19. Sep 9, 2006

0rthodontist

So can you guess what happens as you increase the exponent? Try a few more like .5^3 or .5^10.

Now you may be able to find that limit.

20. Sep 9, 2006

sammieh

O.K. I think I finally got it.

Sn = 4, u1=1 and r=3x/2, so

4 = 1/(1-3x/2)

solve for x and x = 1/2!

FINALLY!