# Convergence help

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

z in C.

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

Thanks.

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vela
Staff Emeritus
Homework Helper

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.

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.
Ratio Test

vela
Staff Emeritus
Homework Helper

You need to explain what you did in more detail.

You need to explain what you did in more detail.
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?

jbunniii
Homework Helper
Gold Member

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?

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|

What does this mean geometrically?
Why did you square it?

jbunniii
Homework Helper
Gold Member

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

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

z + 1 times it conjugate is

$$z^2 - 1$$

jbunniii
Homework Helper
Gold Member

I don't understand what you mean.

z + 1 times it conjugate is

$$z^2 - 1$$
No, that's z + 1 times z - 1. The conjugate of z + 1 is not z - 1.

No, that's z + 1 times z - 1. The conjugate of z + 1 is not z - 1.
So am I supposed to get $\bar{z}> -1$??

jbunniii
Homework Helper
Gold Member

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

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

jbunniii
Homework Helper
Gold Member

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