Values for x which converge. (geometric series)

[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