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[SOLVED] What z makes this converge

dwsmith

Well-known member
Feb 1, 2012
1,673
$\displaystyle\sum_{n=0}^{\infty}\left(\frac{z}{z+1}\right)^n$


$z\in\mathbb{C}$

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?
 
Last edited:

AlexYoucis

New member
Jan 26, 2012
19
$\displaystyle\sum_{n=0}^{\infty}\left(\frac{z}{z+1}\right)^n$


$z\in\mathbb{C}$

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
Feb 1, 2012
1,673

AlexYoucis

New member
Jan 26, 2012
19
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
Feb 1, 2012
1,673
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
Jan 26, 2012
19

dwsmith

Well-known member
Feb 1, 2012
1,673

AlexYoucis

New member
Jan 26, 2012
19
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673

AlexYoucis

New member
Jan 26, 2012
19
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!