I have a simple question. Let $ S = \sum\limits_{k = 1}^n {e^{ikx} } $ using the typical trick , we also have $ S\left( {e^{ix} - 1} \right) = e^{i\left( {n + 1} \right)x} - e^{ix} $ and if $ \left( {e^{ix} - 1} \right) \ne 0 $ we divide by it and we have: $ S=\left( {1 - e^{inx} } \right)\frac{{e^{ix} }} {{1 - e^{ix} }} $ And I see that this limit if n goes to infinity does not exist, but Wolfram|Alpha says that it converges. The problem clearly is in the factor $ \left( {1 - e^{inx} } \right) $ it goes to 0 :S? or something
a question about the sum of $ e^{ikx}$
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calculus
sequences-and-series
wolfram-alpha
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0You have a geometric series with common factor $e^{ix}$. You also (should!) know that $|e^{ix}|=1$; would you happen to remember how the geometric series behaves at the boundary of the region where it converges? – 2011-11-17
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0You are saying that $ \mathop {\lim }\limits_{n \to \infty } e^{ixn} = 0 $ ? =S!! – 2011-11-17
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0I haven't said anything about convergence (yet). What do you remember about the behavior of the geometric series at the boundary? – 2011-11-17
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0the geometric series converges iff the norm of the term is less than 1. – 2011-11-17
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0Yes. But since $|e^{ix}|=1$ for all $x\in \mathbb R$, then... – 2011-11-17
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0But Im not computing the sum of the absolute values :S I´m confused – 2011-11-17
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0let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/1804/discussion-between-susuk-and-j-m) – 2011-11-17
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0What do you mean by norm, then, if not the absolute value? – 2011-11-17