Values for x which converge. (geometric series)

[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