Values for x which converge. (geometric series)

[Jbreezy]

Homework Statement

Find the values of x for which the series converges. Find the sum of the series for those values of x.

Homework Equations

Sum (x+2)^n from n =1 to ∞

The Attempt at a Solution

OK so this is maybe fudging it.
I said that the argument x+2 will diverge if the argument is >= 1. So I just messed with this.

x+2 > = 1 ---> x >= -1 Values of x will diverge.

Then to find the sum this is where it get shady I don't know what to do . Or if the above is OK.

 

[Mark44]

Homework Statement

Find the values of x for which the series converges. Find the sum of the series for those values of x.

Homework Equations

Sum (x+2)^n from n =1 to ∞

The Attempt at a Solution

OK so this is maybe fudging it.
I said that the argument x+2 will diverge if the argument is >= 1. So I just messed with this.

x+2 > = 1 ---> x >= -1 Values of x will diverge.

Then to find the sum this is where it get shady I don't know what to do . Or if the above is OK.

What about if x ≤ -3? Will your series converge or diverge?

 

[Jbreezy]

It will be ((-3)+2)^n = (-1)^n Diverge. Wouldn't it just oscillate between -1 and 1 depending on if n was even or odd?

 

[Jbreezy]

Wait I kind of read that wrong. I think it will converge when x < -3 except that you can't have x = -2?

 

[Mark44]

It will be ((-3)+2)^n = (-1)^n Diverge. Wouldn't it just oscillate between -1 and 1 depending on if n was even or odd?

Yes, the series diverges. The terms of the series oscillate as you said. Using "it" to explain a concept that involves three things (series, sequence of partial sums, sequence of terms) does not lead to clarity.

Wait I kind of read that wrong. I think it will converge when x < -3 except that you can't have x = -2?

Write a few terms of your series when, say, x = -4. What does that series do?

Why can't you have x = -2? What does the series look like in this case?

 

[Jbreezy]

Wait. Wait.

|x+2| = x+2 >= 0 or -(x+2) x< 0 so

x+2 < 1 for convergence so x < -1
and -(x+2) < 1 implies x < - 3

So it converges for -1 < x < -3 ?

So for the sum then what do I do. If I just pretend that x is anything else and do the typical summation of geometric series a/ (1 - r) with a = (x+2) r = (x+2 )

I get S = (x+2)/(-x-1) so do I just specify -1 < x < -3 ?

 

[Mark44]

A lot of what you wrote here makes no sense.

Wait. Wait.

|x+2| = x+2 >= 0 or -(x+2) x< 0 so

No, that's not how the absolute value works.
|x + 2| = x + 2 if x + 2 ≥ 0, and
|x + 2| = -(x + 2) if x + 2 < 0.

x+2 < 1 for convergence so x < -1
and -(x+2) < 1 implies x < - 3

So it converges for -1 < x < -3 ?

I think I know what you're trying to say, but you're not writing it correctly. -1 is not less than -3. It should be the series converges when -3 < x < -1.

So for the sum then what do I do. If I just pretend that x is anything else and do the typical summation of geometric series a/ (1 - r) with a = (x+2) r = (x+2 )

What do you mean "pretend that x is anything else"? The series converges for -3 < x < -1. If x is outside this interval, the series diverges, so why are you trying to find its sum?

I get S = (x+2)/(-x-1) so do I just specify -1 < x < -3 ?

 

[Jbreezy]

A lot of what you wrote here makes no sense.
No, that's not how the absolute value works.
|x + 2| = x + 2 if x + 2 ≥ 0, and
|x + 2| = -(x + 2) if x + 2 < 0.

I thought this was what wiki said.
If |x| = x if x>= 0 and -x if x<0 ?

I think I know what you're trying to say, but you're not writing it correctly. -1 is not less than -3. It should be the series converges when -3 < x < -1.
What do you mean "pretend that x is anything else"? The series converges for -3 < x < -1. If x is outside this interval, the series diverges, so why are you trying to find its sum?

But what about the value for x = -2? Also I'm trying to find the sum because they books say " Find the sum of the series for those values of x.
So that is what I was trying.

 

[Mark44]

I thought this was what wiki said.
If |x| = x if x>= 0 and -x if x<0 ?

Right, but that's not what you wrote, which I've copied below.

|x+2| = x+2 >= 0 or -(x+2) x< 0

You must have been in a hurry, because you left out a lot in what you wrote.

But what about the value for x = -2? Also I'm trying to find the sum because they books say " Find the sum of the series for those values of x.
So that is what I was trying.

What does the series look like if x = -2? I asked this question several posts back (post #5), but you didn't respond.

 

[Jbreezy]

Sorry. I need to slow it down. I have 1000000000 things going on right now. OK so I'm not sure I understand what it would look like. It would be (0)^n = 0. I don't know what it looks like though .. Geometrically ?

 

[Mark44]

Write four or five terms of the series when x = -2. That's what I meant by "what does it look like?"

 

[Jbreezy]

Wouldn't it just be 0?

(x+2)^n = (-2 +2) ^n + (-2 +2)^n + (-2+2)^n ?

But what if X was just a tad bit less then -2 maybe -2.1 or -2.3. or in the other direction. I'm not sure what I'm supposed to think here. It gets smaller and smaller and smaller as it goes closer to two.

Maybe its like -3 < x < -1 (x not equal to two).
What am I supposed to do about the sum. Is the sum two then ? Because that is what it approaches?

 

[Mark44]

Wouldn't it just be 0?

(x+2)^n = (-2 +2) ^n + (-2 +2)^n + (-2+2)^n ?

?

Your series is
$$\sum_{n = 1}^{\infty}(x + 2)^n$$
Write the first five terms of this series when x = -2.

But what if X was just a tad bit less then -2 maybe -2.1 or -2.3. or in the other direction. I'm not sure what I'm supposed to think here. It gets smaller and smaller and smaller as it goes closer to two.

Please don't use the word "it", especially when "it" refers to two different things. Not only is that confusing to someone who is trying to understand what you're saying, but it also shows that things are very unclear in your mind. This is like saying, "He's coming over tonight, but I'll see him tomorrow" when the first "he" is Bob, but the "him" is Bill.

Maybe its like -3 < x < -1 (x not equal to two).
What am I supposed to do about the sum. Is the sum two then ? Because that is what it approaches?

Things aren't gelling here for you. You're still asking questions about things that you have already answered.

What is the interval of convergence? You've already said that the series converges for -3 < x < -1, right? If x is any number in that interval, then the series converges. Period.

The series is a geometric series, right? How do you find the sum of such a series for a particular value of x? You have a formula, right?

What if the formula doesn't work for a particular x value, especially x = -2? You try something else, such as writing a few terms in the series. If you do this, you should immediately see what happens.

 

[Jbreezy]

$\sum_{n = 1}^{\infty}((-2) + 2)^n = (0)^1 + (0)^2 + (0)^3 + ...$

 

[Jbreezy]

$\sum_{n = 1}^{\infty}((-2) + 2)^n = (0)^1 + (0)^2 + (0)^3 + ...$

I'm sorry I posted that before I was ready ignore my previous one.

I don't understand. When x = -2 it is 0. I don't get what is the deal.

I'm asking ...for the sum do I just use the formula and treat x as a number?

$\sum_{n = 1}^{\infty}(x + 2)^n = \sum_{n = 1}^{\infty}(x + 2)(x + 2)^{n-1} = (x-2)/( 1-(x +2)) = (x-2)/ (-x-1)$ Then do I just specify the interval ?

Like -3 < x < -1 ...now this leads to my question about what to do when x = 2? Which brings us back to the top of this post.

I'm sorry I'm trying here I don't understand this.

 

[Ray Vickson]

$\sum_{n = 1}^{\infty}((-2) + 2)^n = (0)^1 + (0)^2 + (0)^3 + ...$

I'm sorry I posted that before I was ready ignore my previous one.

I don't understand. When x = -2 it is 0. I don't get what is the deal.

I'm asking ...for the sum do I just use the formula and treat x as a number?

$\sum_{n = 1}^{\infty}(x + 2)^n = \sum_{n = 1}^{\infty}(x + 2)(x + 2)^{n-1} = (x-2)/( 1-(x +2)) = (x-2)/ (-x-1)$ Then do I just specify the interval ?

Like -3 < x < -1 ...now this leads to my question about what to do when x = 2? Which brings us back to the top of this post.

I'm sorry I'm trying here I don't understand this.

What are you not seeing? You have already told us that the interval of convergence is -3 < x < -1. The value x = 2 lies way outside the interval of convergence, so what does that tell you? The value x = -2 lies inside the interval of convergence, so what does that tell you?

As for using the summation formula and then worrying about the interval: that is going about it backwards. FIRST decide on the interval of convergence, and THEN find a formula for the sum. For x outside the interval of convergence, the final sum formula does NOT represent the sum because the sum does not exist, but the formula does.

 

[Jbreezy]

What are you not seeing? You have already told us that the interval of convergence is -3 < x < -1. The value x = 2 lies way outside the interval of convergence, so what does that tell you? The value x = -2 lies inside the interval of convergence, so what does that tell you?

As for using the summation formula and then worrying about the interval: that is going about it backwards. FIRST decide on the interval of convergence, and THEN find a formula for the sum. For x outside the interval of convergence, the final sum formula does NOT represent the sum because the sum does not exist, but the formula does.

It tells me x = 2 diverges because it is outside and it converges for x = -2. But when you evaluate it for x = -2 you just get 0 for the sum. Converges to 0?

So is my formula correct and I just need to specify the values for which it works along side it?

 

[Dick]

It tells me x = 2 diverges because it is outside and it converges for x = -2. But when you evaluate it for x = -2 you just get 0 for the sum. Converges to 0?

So is my formula correct and I just need to specify the values for which it works along side it?

The formula you quoted in post #15 is NOT correct. Why did you change the (x+2) you factored out into (x-2)?

 

[Mark44]

$\sum_{n = 1}^{\infty}((-2) + 2)^n = (0)^1 + (0)^2 + (0)^3 + ...$

= 0, clearly. Was that so hard?

I'm sorry I posted that before I was ready ignore my previous one.

I don't understand. When x = -2 it is 0. I don't get what is the deal.

I'm asking ...for the sum do I just use the formula and treat x as a number?

x is a number, but the formula below is incorrect. It should give 0 when x = -2, but it doesn't.

$\sum_{n = 1}^{\infty}(x + 2)^n = \sum_{n = 1}^{\infty}(x + 2)(x + 2)^{n-1} = (x-2)/( 1-(x +2)) = (x-2)/ (-x-1)$ Then do I just specify the interval ?

Like -3 < x < -1 ...now this leads to my question about what to do when x = 2? Which brings us back to the top of this post.

I'm sorry I'm trying here I don't understand this.

 

[Mark44]

It tells me x = 2 diverges because it is outside and it converges for x = -2. But when you evaluate it for x = -2 you just get 0 for the sum.

x = 2 neither converges nor diverges. x = 2 is just a number.

Converges to 0?

What does? PLEASE DON'T SAY "IT"!

 

[Jbreezy]

The formula you quoted in post #15 is NOT correct. Why did you change the (x+2) you factored out into (x-2)?

Not sure. Error. If I fix that then it is OK.

x = 2 neither converges nor diverges. x = 2 is just a number.

Converges to 0?

What does? PLEASE DON'T SAY "IT"!

I just meant that when x = 2 then the formula gives 0. So I was asking if it is converging to 0? Maybe this is the same as when x = 2 the formula just gives a number?

 

[Dick]

Not sure. Error. If I fix that then it is OK.

I just meant that when x = 2 then the formula gives 0. So I was asking if it is converging to 0? Maybe this is the same as when x = 2 the formula just gives a number?

If you fixed the formula then when x=2 it doesn't give 0. But that doesn't really matter. The formula is only valid for -3<x<-1. What it gives outside of that range has nothing to do with anything.

 

[Jbreezy]

It should still give 0 though. RIght?

 

[Mark44]

It should still give 0 though.

It = who?

RIght?

 

[Jbreezy]

God dang it. It = the formula I cam up with for the sum.

 

[Mark44]

God dang it. It = the formula I cam up with for the sum.

Which formula? As has already been pointed out by Dick and me, the formula you showed in post #15 is wrong.

 

[Mark44]

Jbreezy, the reason I've been harping at you to stop using "it" is that you are not communicating clearly. If I said this to you -- "I used to have a dog and a cat, but it died." -- would you know which animal I was talking about?

 

[Jbreezy]

No I wouldn't know. I know my language sucks.

Did I just make a mistake with the formuala the algebra just go about it wrong? Or is the concept wrong?

 

[Mark44]

No I wouldn't know.

That's exactly my point.

I know my language sucks.

You can improve it by not using unclear language.

Did I just make a mistake with the formuala the algebra just go about it wrong? Or is the concept wrong?

Show us the formula you're talking about.

 

[Jbreezy]

$\sum_{n = 1}^{\infty}(x + 2)^n = \sum_{n = 1}^{\infty}(x + 2)(x + 2)^{n-1} = (x-2)/( 1-(x +2)) = (x-2)/ (-x-1)$
This formula

 

[Mark44]

Please pay attention. Dick told you about a dozen posts ago that your formula in post #15 was incorrect.

The formula you quoted in post #15 is NOT correct. Why did you change the (x+2) you factored out into (x-2)?

$\sum_{n = 1}^{\infty}(x + 2)^n = \sum_{n = 1}^{\infty}(x + 2)(x + 2)^{n-1} = (x-2)/( 1-(x +2)) = (x-2)/ (-x-1)$
This formula

 

[Jbreezy]

Yeah I know the formula is incorrect. I'm asking if it is wrong all together or if I just made a mistake in the algebra! Thats all I'm asking.

 

[Mark44]

$\sum_{n = 1}^{\infty}(x + 2)^n = \sum_{n = 1}^{\infty}(x + 2)(x + 2)^{n-1} = (x-2)/( 1-(x +2)) = (x-2)/ (-x-1)$
This formula

Yeah I know the formula is incorrect. I'm asking if it is wrong all together or if I just made a mistake in the algebra! Thats all I'm asking.

Well, this part is OK, but pointless.
$\sum_{n = 1}^{\infty}(x + 2)^n = \sum_{n = 1}^{\infty}(x + 2)(x + 2)^{n-1}$
As I said in the other thread, $\sum_{n = 1}^{\infty} r^n = \frac r {1 - r}$

In your work above, the parts after the 2nd = are wrong. We've been waiting for you to fix them since post #15, which is 18 posts ago.

 

[Jbreezy]

I'm sorry I feel like you summation is wrong. No one write it as r/1-r
In my book it is A! Where a is the coefficient of r. And they do the formula to r^(n-1) when it starts at n = 1

 

[Mark44]

I'm sorry I feel like you summation is wrong.

Nope. I derived how I got it for that summation ($\sum_{n = 1}^{\infty}r^n$) in the other thread.

If the series is $\sum_{n = 0}^{\infty}r^n$ I get a different value, 1/(1 - r), again assuming that |r| < 1.

No one write it as r/1-r

Correct. They would write it as r/(1 - r).

In my book it is A!

Who is it?

Where a is the coefficient of r. And they do the formula to r^(n-1) when it starts at n = 1

 

[Mark44]

Let's stay focused here. How can you fix the equation you posted in post #15?

$\sum_{n = 1}^{\infty}(x + 2)^n = \sum_{n = 1}^{\infty}(x + 2)(x + 2)^{n-1} = (x-2)/( 1-(x +2)) = (x-2)/ (-x-1)$

This is the part that's wrong:
(x-2)/( 1-(x +2)) = (x-2)/ (-x-1)

 

[Jbreezy]

Let's stay focused here. How can you fix the equation you posted in post #15?


This is the part that's wrong:
(x-2)/( 1-(x +2)) = (x-2)/ (-x-1)

That algebra is OK though. Right. I don't see an error int at so maybe something else.

 

[Mark44]

Here's what's wrong.
$$\sum_{n = 1}^{\infty}(x + 2)^n \neq (x-2)/( 1-(x +2)) = (x-2)/ (-x-1)$$

The expression in the middle is equal to the expression on the right, but the series on the left does NOT add up to what you have in the middle.

This is what we've been saying since you posted the above way back in post #15.

 

[Jbreezy]

$\sum_{n = 1}^{\infty}(x+2)(x + 2)^{n-1} \neq (x-2)/( 1-(x +2)) = (x-2)/ (-x-1)$

I know you don't like it

I don't know how to get rid of the not equal

 

[Mark44]

Right, I don't like it. Why don't you write it like this?
$$ \sum_{n = 1}^{\infty}(x + 2)^2(x + 2)^{n -2}$$
Or maybe like this?
$$ \sum_{n = 1}^{\infty}(x + 2)^3(x + 2)^{n -3}$$
These aren't wrong, but they're silly. What I'm saying is that you are needlessly complicating things, which is never a good idea.

Here's your series, in unsilly form:
$$ \sum_{n = 1}^{\infty}(x + 2)^n$$

This is what I said in the other thread you started today. It definitely applies to the series you have in this thread.

For a geometric series such as $\sum_{n = 1}^{\infty}r^n$, if |r| < 1, the series converges to r/(1 - r).

Here's a simplistic explanation:
Let S = r + r² + ... + rⁿ + ...
Then -rS = -r² - r³ - ... - r^{n + 1} - ...

Then S - rS = r, or S(1 - r) = r, so S = r/(1 - r)

 

[Jbreezy]

$\sum_{n = 1}^{\infty}(x + 2)^n$ = (x+2)/(1-(x+2))

r = (x+2) right? no..idk I think it's ok?

 

[Mark44]

YES!

So when x = -2, which is a value in the interval of convergence, what is the sum of the series?

 

[Jbreezy]

$IT$ would be 0 jajaj