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 &lt; 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 &lt; 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&lt;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