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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.
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.

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.1This is my attempt: if z =1, the sum is \(\sum_{n=1}^\infty n^a\) converges if and only if a = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That can't be right, \(\sum\limits_{n\ge1}\frac{1}{n}\) diverges...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think you meant to say \(a<1\)?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1oh, yeah, I thought it diverges. if z = 1, the sum is \(\sum_{n=1}^\infty n^a (1)^n\) converges for a <0, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Hence combine the two, only a =1 satisfies both, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Condition 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?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1or \(1\leq a<0\)?? can it be?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I agree with your conclusion for the second series, but for \(z=1\) the series diverges for \(a\ge1\). Right, it should be \(1\le a<0\).

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1But on this interval, discuss the convergence of the sum is not easy (part b). I don't know how to argue.:(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Any 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...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For the series to converge, both the real and imaginary parts must also converge. Notice that \[n^a\cos nt=n^a\cos nt\le n^a~~\implies~~\sum_{n=1}^\infty n^a\cos nt\le\sum_{n=1}^\infty n^a\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The same can be said for the imaginary part.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Thank you. I got it. :)
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