# How does a geometric series converge, or have a sum?

1. Apr 21, 2015

### RyanTAsher

1. The problem statement, all variables and given/known data

How does a geometric series have a sum, or converge?

2. Relevant equations

Sum of Geometric Series = $\frac {a} {1-r}$

If r ≥ ±1, the series diverges. If -1 < r < 1, the series converges.

3. The attempt at a solution

How exactly does a infinite geometric series have a sum, or converge (tend to) a specific limit?

I understand that it is due to partial sums that we are able to derive the formula for the sum of a geometric series, yet at the same time I don't understand how a sequence that will be always multiplied by itself to infinity can ever STOP and have a final sum, or how it converges (tends toward) a specific limit.

For example $\sum\limits_{k=1}^{∞} (\frac {2} {3})^k$. This sum works out to 3, and does converge as r > -1 and r < 1. Why??? The partial sums proof makes no sense to me, and how on earth do things ever cancel to end up summing to 3?

Please explain to me if I have some conceptual misunderstanding of converges or sums of series, or if I'm just overlooking something...

2. Apr 21, 2015

### SammyS

Staff Emeritus
What proof are you familiar with?

What is it in that proof that you don't understand, or what is it that you can't 'buy'?

3. Apr 21, 2015

### kiritee Gak

Converging to something means it doesn't matter how much to add to that series(by next terms) (since they get smaller and smaller in our case) the sum does not exceed a particular value(the converging point).what is it u specifically want?? cheers.

4. Apr 21, 2015

### RyanTAsher

I don't exactly understand how at the end of the proof you can just ignore all the terms in between the kth terms and first few non-kth terms, and just divide those out. I understand the factoring out the a of the general equation ar^k, but how can you ignore the quotient of all the in between terms.

5. Apr 21, 2015

### RyanTAsher

That actually helps a lot to think of it that way... In other words since r ≤ ±1, the terms become so insignificant that their effect on the sum is negligible?

6. Apr 21, 2015

### SammyS

Staff Emeritus
For a series in general, having successive terms get smaller does not guarantee convergence of the series.

7. Apr 21, 2015

### kiritee Gak

i think i specifically mentioned "in our case" which is 2/3 series Ryan talking.

8. Apr 21, 2015

### Staff: Mentor

Please don't use "text speak" ("u" for "you") here at PF. From the forum rules (https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/)

9. Apr 21, 2015

### Staff: Mentor

Their effect is not insignificant, but they don't cause the partial sums to go past a specific value. Here's an example of a geometric series with r = 0.9.
$$\sum_{n = 1}^{\infty} \frac 9 {10^n}$$
The partial sums go like this:
S1 = .9
S2 = .9 + .09 = .99
S3 = .9 + .09 + .009 = .999
and so forth.
Each partial sum merely adds one more 9 onto the end of the decimal fraction. All of the partial sums are bounded above by 1, but successive partial sums get closer and closer to 1 as more terms are added, allowing us to conclude that the sum of the series is 1.

10. Apr 23, 2015

### Fredrik

Staff Emeritus
The sum is 3 because for each n, the nth partial sum is
$$\frac{1-\big(\frac 2 3\big)^{n+1}}{1-\frac 2 3} = 3-3\left(\frac 2 3\right)^{n+1}$$ and the second term on the right goes to zero as n goes to infinity. If that doesn't make any sense to you, there's probably something missing in your understanding of limits of sequences. Perhaps you can be more specific about what's bothering you.

Note in particular that every partial sum is less than 3.

11. Apr 23, 2015

### RyanTAsher

Thank you that makes a lot more sense when you use the partial sums equation, rather than just the sum of an infinite geometric series equation. It makes sense now!

12. Apr 24, 2015

### epenguin

Your problem seemed to be you could not argue with the algebra yet could not imagine how it could be. Attached (excuse crudity of my first attempt with whiteboard but I hope it gives the idea) a visualisation convincingly showing
1/2 + 1/4 + 1/8 + ... → 1

View attachment 82466

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• ###### Screenshot 2015-04-24 09.55.50.png
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Last edited: Apr 24, 2015
13. Apr 24, 2015

### RyanTAsher

Thank you for the help, the visual representation helps a lot more too, actually being able to see something, essentially converge.

14. Apr 25, 2015