# [SOLVED]What z makes this converge

#### dwsmith

##### Well-known member
$\displaystyle\sum_{n=0}^{\infty}\left(\frac{z}{z+1}\right)^n$

$z\in\mathbb{C}$

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?

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#### AlexYoucis

##### New member
$\displaystyle\sum_{n=0}^{\infty}\left(\frac{z}{z+1}\right)^n$

$z\in\mathbb{C}$

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?

#### dwsmith

##### Well-known member
You solve that inequality..surely you can do that, no?
Apparently I can't because I keep getting it wrong.

#### AlexYoucis

##### New member
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 $x>\frac{-1}{2}$.

#### dwsmith

##### Well-known member
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

##### New member
How did you go from this $|z|<|z+1|$ to this $x^2+y^2<(x+1)^2+y^2$??
Let $z=x+iy$.

#### AlexYoucis

##### New member
I did but I don't see what happened to all the i's.
If $z=x+iy$ then $|z|=\sqrt{x^2+y^2}$.

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#### AlexYoucis

##### New member
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!