# Finding the values of $z$ s.t. $\sum_{n=0}^{\infty} \left( \frac{z}{1+z} \right)^n$ is convergent

I manipulated the series to $\sum_{n=0}^{\infty} \left( \frac{1}{1-(-1/z)} \right)^n$, which converges for $|-1/z|<1$ by geometric series. Then solving for $z$, I obtained $z>(1/\bar{z})$.

Is this correct? I was expecting some numerical values, so I'm not sure about my answer.

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## 1 Answer

Let me suggest another approach. Put $w=\dfrac{z}{z+1}$. Where will $$\sum_{n=0}^\infty w^n$$ converge? What does this tell you about the possible domain of $z$?

ADD

I am getting $$2\Re z+1>0$$ Note you cannot really use complex numbers in an inequality, since they are not ordered in any way. However, from $|z|<|1+z|$ you get the equivalent $$z\bar z<(1+z)(1+\bar z)$$ since everything is real. This gives $$z\bar z <1+(z+\bar z)+z\bar z$$ which is the same as $$0<1+2\Re (z)$$

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$z > 1-\bar{z}$? –  AlanH May 26 '13 at 22:36
@AlanH I added something. –  Pedro Tamaroff May 26 '13 at 22:40
yeah, I got the same thing, but I just didn't express the final step like that. Does it matter though? –  AlanH May 26 '13 at 22:46
@AlanH What matters is that complex numbers aren't ordered, so $<$ makes no sense for complex numbers, so you ought to express $|z|<|z+1|$ in terms of the real and imaginary parts of $z$. –  Pedro Tamaroff May 26 '13 at 22:47