The series $\displaystyle\sum_{n=0}^{\infty}\left(\frac{z}{z+1}\right)^n$ converges when the condition $\left|\frac{z}{z+1}\right|<1$ is satisfied. This inequality can be solved by transforming it into the form $|z|<|z+1|$, leading to the conclusion that $x>\frac{-1}{2}$, where $z=x+iy$. The discussion emphasizes the importance of correctly applying the ratio test and understanding the implications of complex numbers in convergence analysis.
PREREQUISITESMathematicians, students of complex analysis, and anyone interested in series convergence and the application of the ratio test in complex variables.
You solve that inequality..surely you can do that, no?dwsmith said:$\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 said:You solve that inequality..surely you can do that, no?
dwsmith said:Apparently I can't because I keep getting it wrong.
AlexYoucis said: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 said:How did you go from this $|z|<|z+1|$ to this $x^2+y^2<(x+1)^2+y^2$??
AlexYoucis said:Let $z=x+iy$.
dwsmith said:I did but I don't see what happened to all the i's.
AlexYoucis said:If $z=x+iy$ then $|z|=x^2+y^2$.
dwsmith said:Shouldn't it be square rooted?