# Convergence help

1. Feb 1, 2012

### fauboca

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

z in C.

This only converges for $z>\frac{-1}{2}$, correct?

Thanks.

2. Feb 1, 2012

### vela

Staff Emeritus
Re: Convergence

No. For one thing, z is complex, so it doesn't make sense to say z > -1/2. In any case, show us how you reached this conclusion.

3. Feb 1, 2012

### fauboca

Re: Convergence

Ratio Test

4. Feb 1, 2012

### vela

Staff Emeritus
Re: Convergence

You need to explain what you did in more detail.

5. Feb 1, 2012

### fauboca

Re: Convergence

By the ratio test,

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

Is this correct now?

If so, how do I find the appropriate z for which it converges?

6. Feb 1, 2012

### jbunniii

Re: Convergence

Your inequality is true if and only if

$$\left|\frac{z}{z+1}\right|^2 < 1$$

You can rearrange this and simplify to find a simple inequality for z.

Another way: your inequality is true if and only if

|z| < |z + 1|

which can be written more suggestively as

|z - 0| < |z + 1|

What does this mean geometrically?

7. Feb 1, 2012

### fauboca

Re: Convergence

Why did you square it?

8. Feb 1, 2012

### jbunniii

Re: Convergence

Because you can then simplify $|z + 1|^2$. If you're not sure how, try writing it as z + 1 times its complex conjugate.

9. Feb 1, 2012

### fauboca

Re: Convergence

I don't understand what you mean.

z + 1 times it conjugate is

$$z^2 - 1$$

10. Feb 1, 2012

### jbunniii

Re: Convergence

No, that's z + 1 times z - 1. The conjugate of z + 1 is not z - 1.

11. Feb 1, 2012

### fauboca

Re: Convergence

So am I supposed to get $\bar{z}> -1$??

12. Feb 1, 2012

### jbunniii

Re: Convergence

No. That doesn't even make any sense, as $\bar{z}$ is a complex number.

$$|z+1|^2 = (z+1)(\overline{z + 1}) = (z + 1)(\bar{z} + 1) = z\bar{z} + z + \bar{z} + 1 = |z|^2 + z + \bar{z} + 1$$

So what does your inequality become?

Hint: you can simplify $z + \bar{z}$.

13. Feb 1, 2012

### fauboca

Re: Convergence

I defined $z = x + yi$ and then did what you said. But we had $|z| < |z+1|$ That is how I obtained my solution.

14. Feb 1, 2012

### jbunniii

Re: Convergence

But you didn't obtain a correct solution, or at least you haven't posted it here.

OK, forget the squaring for now and let's focus on this inequality:

|z| < |z + 1|

What does this mean geometrically?

I will write it more suggestively as follows:

|z - 0| < |z - (-1)|

What values of z satisfy this inequality? It's a certain region of the complex plane. What region?

This kind of thing is often easier to answer if you first determine what region is specified by the EQUALITY:

|z - 0| = |z - (-1)|