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

[Destroxia]

Homework Statement

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

Homework Equations

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

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

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...

 

[SammyS]

Homework Statement

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

Homework Equations

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

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

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...

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'?

 

[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.

 

[Destroxia]

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'?

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.

 

[Destroxia]

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.

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?

 

[SammyS]

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 you specifically want?? cheers.

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

 

[kiritee Gak]

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

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

 

[Mark44]

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.

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/)

SMS messaging shorthand ("text-message-speak"), such as using "u" for "you", "please" for "please", or "wanna" for "want to" is not acceptable.

 

[Mark44]

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?

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.

 

[Fredrik]

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...

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.

 

[Destroxia]

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.

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!

 

[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

https://www.physicsforums.com/attachments/82466

 

[Destroxia]

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

https://www.physicsforums.com/attachments/82466

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

 

[jbunniii]

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

More visual aids for geometric series: https://sites.google.com/site/butwhymath/m/geometric-series-visually