## Loser66 one year ago $$\sum_{n=1}^\infty n^az^n$$ 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.

1. Loser66

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

2. anonymous

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

3. anonymous

I think you meant to say $$a<-1$$?

4. Loser66

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?

5. Loser66

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

6. Loser66

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

7. Loser66

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

8. anonymous

I agree with your conclusion for the second series, but for $$z=1$$ the series diverges for $$a\ge1$$. Right, it should be $$-1\le a<0$$.

9. Loser66

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

10. Loser66

@chris00

11. anonymous

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...

12. anonymous

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

13. anonymous

The same can be said for the imaginary part.

14. Loser66

Thank you. I got it. :)