MHB What Values of z Make This Series Converge?

[Dustinsfl]

$\displaystyle\sum_{n=0}^{\infty}\left(\frac{z}{z+1}\right)^n$$z\in\mathbb{C}$

By the ratio test,

$\displaystyle\left|\frac{z}{z+1}\right|<1$

I am stuck at this part. How do I find the z such that some is convergence?

 

[AlexYoucis]

$\displaystyle\sum_{n=0}^{\infty}\left(\frac{z}{z+1}\right)^n$$z\in\mathbb{C}$

By the ratio test,

$\displaystyle\left|\frac{z}{z+1}\right|<1$

I am stuck at this part. How do I find the z such that some is convergence?

You solve that inequality..surely you can do that, no?

 

[Dustinsfl]

You solve that inequality..surely you can do that, no?

Apparently I can't because I keep getting it wrong.

 

[AlexYoucis]

Apparently I can't because I keep getting it wrong.

Ok, so we need to solve $|z|<|z+1|$ or $x^2+y^2<(x+1)^2+y^2$ or $x^2<(x+1)^2$ or $x>\frac{-1}{2}$.

 

[Dustinsfl]

Ok, so we need to solve $|z|<|z+1|$ or $x^2+y^2<(x+1)^2+y^2$ or $x^2<(x+1)^2$ or $x>\frac{-1}{2}$.

How did you go from this $|z|<|z+1|$ to this $x^2+y^2<(x+1)^2+y^2$??

 

[AlexYoucis]

How did you go from this $|z|<|z+1|$ to this $x^2+y^2<(x+1)^2+y^2$??

Let $z=x+iy$.

 

[Dustinsfl]

Let $z=x+iy$.

I did but I don't see what happened to all the i's.

 

[AlexYoucis]

I did but I don't see what happened to all the i's.

If $z=x+iy$ then $|z|=\sqrt{x^2+y^2}$.

 

[Dustinsfl]

If $z=x+iy$ then $|z|=x^2+y^2$.

Shouldn't it be square rooted?

 

[AlexYoucis]

Shouldn't it be square rooted?

Yes, it should have been, and then I proceeded from there. Do you see how?

P.S. Sorry if I sound terse, I'm just busy doing other things!