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 February 19th, 2013, 01:22 PM #1 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 series convergence? Hi i am trying to figure out how this series would behave: $\sum_{n=-2}^\infty n (\alpha)^n e^{i \omega n }$ given that $|\alpha|<1$ $=\sum_{n=-2}^\infty n (\alpha e^{i \omega })^n$ can anyone give me a hand?
 February 19th, 2013, 01:25 PM #2 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: series convergence? Ignoring the first few terms , doesn't that seem like the derivative of a geometric series ?
 February 19th, 2013, 01:41 PM #3 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: series convergence? kind but I dont think changing the index will help because of the 'n' I think the geometric series formulas dont apply
 February 19th, 2013, 11:20 PM #4 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: series convergence? You want to know the convergence, right? $\sum_{n= - 2}^{\infty} n \cdot \alpha^n e^{i \omega n} = \sum_{n = -2}^{\infty} n \cdot \alpha^n \cos(\omega n) + \mathbb{i} \cdot \sum_{n = -2}^{\infty} n \cdot \alpha^n \sin(\omega n)$ Now, since both sin and cos are bounded by 1, the imaginary, and the real part can be bounded by the sum $\sum_{n=-2}^{\infty} n \cdot \alpha^n$. Since we are given that $| \alpha |$ is smaller than 1, the sum above converges by ratio test. Hence, the original series converges. You said you are interested in the behavior, do you meant that you want to know the growth? Challenge : Find an analytic formula for this series!
 February 20th, 2013, 04:29 AM #5 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: series convergence? By comparison test : $|na^ne^{n\omega i}|\leq |na^n| \,\,$ The sum is absolutely convergent since the latter is a derivative of a geometric series ...

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