Does \sum_{n=1}^{\infty} \frac{z^n}{n} Converge for |z| \leq 1, z \neq 1?

[latentcorpse]

How does one show that $\sum_{n=1}^{\infty} \frac{z^n}{n}$ converges $\forall z s.t. |z| \leq 1, z \neq 1$

i can show that it doesn't converge for z=1 (easy enough) but how do i do the rest of it?

 

[Billy Bob]

Try ratio test and Abel's test.

 

[latentcorpse]

ratio test gives $\frac{z}{1+\frac{1}{n}} \rightarrow z$ as n goes to infinity

so this will be convergent for all z with |z|<1

i don't really understand abels test - how does that help here?

also it says you can't apply abels test on the boundary so how do i show it converges there also?

 

[Billy Bob]

http://en.wikipedia.org/wiki/Abel's_test

Abel's works on the boundary |z|=1 except at z=1. Are you allowed to use Abel's test? If not, I guess you'd have to wade through its proof? The key step is summation by parts, along with a sine trick. I'm assuming z is complex. Real case is easy.

 

[latentcorpse]

ok so abels test can be used here o say that it converges on the boundary ASSUMING that we know it converges for all |z|<1 - and we know this from the ratio test.

but abels test doesn't work on z=1 but we can tell it doesn't converge there.
and then we are done.

is that all acceptable?

 

[n!kofeyn]

How does one show that $\sum_{n=1}^{\infty} \frac{z^n}{n}$ converges $\forall z s.t. |z| \leq 1, z \neq 1$

i can show that it doesn't converge for z=1 (easy enough) but how do i do the rest of it?

There should be a theorem or proposition in your book that says:
If $\sum a_n (z-a)^n$ is a given power series with radius of convergence $R$, then
$$R = \lim_{n\to \infty} \left| \frac{a_n}{a_{n+1}} \right|$$
if this limit exists.

 

[latentcorpse]

wouldn't that limit be 0 as $z^n < z^{n+1} \Rightarrow \lim_{n \rightarrow \infty} |\frac{z^n}{z^{n+1}}|=0$ as the n bits just go to 1 in the limit

 

[n!kofeyn]

wouldn't that limit be 0 as $z^n < z^{n+1} \Rightarrow \lim_{n \rightarrow \infty} |\frac{z^n}{z^{n+1}}|=0$ as the n bits just go to 1 in the limit

Look what I posted. I didn't say $z^n$. The formula I gave has $a_n$'s in it.

 

[Billy Bob]

Thm 1: If $$\left|\frac{b_{n+1}}{b_n}\right|\to L<1$$ then $$\sum b_n$$ converges absolutely.

Thm 2: If $$\left|\frac{a_{n+1}}{a_n}\right|\to L$$ then $$R=1/L$$ is the radius of convergence of $$\sum a_n z^n$$.

You can use either thm. Which one do you want to use? If you want to use Thm 1, you have to use $$b_n=a_n z^n$$. If Thm 2, just use $$a_n$$