Question

\(\sum_{n=1}^\infty n^az^n\) for |z| =1, \(z\neq -1\)
a) Which value of a , the series diverges at z =1 and converges at z = -1?
b) discuss the convergence of the series with the values of a above when |z|=1 \(z\neq 1\)
Please, help.

Answer 1

This is my attempt:
if z =1, the sum is \(\sum_{n=1}^\infty n^a\) converges if and only if a = -1

Answer 2

That can't be right, \(\sum\limits_{n\ge1}\frac{1}{n}\) diverges...

Answer 3

I think you meant to say \(a<-1\)?

Answer 4

oh, yeah, I thought it diverges.
if z = -1, the sum is \(\sum_{n=1}^\infty n^a (-1)^n\) converges for a <0, right?

Answer 5

Hence combine the two, only a =-1 satisfies both, right?

Answer 6

Condition for z = 1 and the sum is divergent is a =-1
condition for z =-1 and the sum is convergent is a <0
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the condition of a which satisfies both is a =-1 , right?

Answer 7

or \(-1\leq a<0\)?? can it be?

Answer 8

I agree with your conclusion for the second series, but for \(z=1\) the series diverges for \(a\ge1\).

Right, it should be \(-1\le a<0\).

Answer 9

But on this interval, discuss the convergence of the sum is not easy (part b). I don't know how to argue.:(

Answer 10

@chris00

Answer 11

Any complex number that lies on the circle \(|z|=1\) with \(z\neq1\) can be written as \(z=e^{it}\), with \(0<t<2\pi\).

So you have
\[\sum_{n=1}^\infty n^az^n=\sum_{n=1}^\infty n^ae^{int}=\sum_{n=1}^\infty n^a\cos nt+i\sum_{n=1}^\infty n^a\sin nt\]That should make things easier to work with...

Answer 12

For the series to converge, both the real and imaginary parts must also converge. Notice that
\[|n^a\cos nt|=n^a|\cos nt|\le n^a~~\implies~~\sum_{n=1}^\infty n^a\cos nt\le\sum_{n=1}^\infty n^a\]

Answer 13

The same can be said for the imaginary part.

Answer 14

Thank you. I got it. :)